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First Published 2016
ISBN 978 981 288 014 7
Printed in Singapore
PREFACE
O Level Additional Mathematics Topical Revision Notes has been written in
accordance with the latest syllabus issued by the Ministry of Education, Singapore.
This book is divided into 16 units, each covering a topic as laid out in the syllabus.
Important concepts and formulae are highlighted in each unit, with relevant worked
examples to help students learn how to apply theoretical knowledge to examination
questions.
To make this book suitable for N(A) Level students, sections not applicable for
the N(A) Level examination are indicated with a bar ( ).
We believe this book will be of great help to teachers teaching the subject and students
preparing for their O Level and N(A) Level Additional Mathematics examinations.
Preface
iii
CONTENTS
Unit 1
Simultaneous Equations, Polynomials and Partial Fractions
Unit 2
Quadratic Equations, Inequalities and Modulus Functions
25
Unit 3
Binomial Theorem
44
Unit 4
Indices, Surds and Logarithms
52
Unit 5
Coordinate Geometry
67
Unit 6
Further Coordinate Geometry
77
Unit 7
Linear Law
85
Unit 8
Trigonometric Functions and Equations
93
Unit 9
Trigonometric Identities and Formulae
104
Unit 10
Proofs in Plane Geometry
113
Unit 11
Differentiation and its Applications
121
Unit 12
Further Applications of Differentiation
129
Unit 13
Differentiation of Trigonometric, Logarithmic &
139
1
Exponential Functions and their Applications
Unit 14
Integration
148
Unit 15
Applications of Integration
158
Unit 16
Kinematics
170
Mathematical Formulae
iv Contents
178
UNIT
1
Simultaneous Equations,
Polynomials and Partial
Fractions
Simultaneous Linear Equations
1.
The solution(s) of a pair of linear and/or non-linear equations correspond to the
coordinates of the intersection point(s) of the graphs.
2. A pair of simultaneous linear equations is of the form
ax + by = p
cx + dy = q,
where
a, b, c and d are constants,
x and y are variables to be determined.
3.
There is usually one solution to a pair of simultaneous linear equations.
4. Methods of solving simultaneous linear equations:
• Elimination (covered in ‘O’ level Mathematics)
• Substitution (covered in ‘O’ level Mathematics)
• Matrix method (not in syllabus)
• Graphical method (covered in ‘O’ level Mathematics)
Simultaneous Equations, Polynomials and Partial Fractions
1
5. The methods most commonly used to solve simultaneous linear equations are
• Elimination
The coefficient of one of the variables is made the same in both equations. The
equations are then either added or subtracted to form a single linear equation
with only one variable.
Example 1
Solve the simultaneous equations
Solution
2x + 3y = 15 ––– (1)
–3y + 4x = 3 ––– (2)
(1) + (2):
(2x + 3y) + (–3y + 4x) = 18
6x = 18
x=3
When x = 3, y = 3.
2
UNIT 1
2x + 3y = 15
–3y + 4x = 3
•
Substitution
A variable is made the subject of the chosen equation. This equation is then
substituted into the equation that was not chosen to solve for the variable.
Example 2
Solve the simultaneous equations
2x – 3y = –2,
y + 4x = 24.
Solution
2x – 3y = –2 ––– (1)
y + 4x = 24 ––– (2)
From (1):
−2 + 3y
x=
2
3
x = –1 + y ––– (3)
2
Substitute (3) into (2):
⎛
3 ⎞
y + 4 ⎜–1 + y⎟ = 24
2 ⎠
⎝
y – 4 + 6y = 24
7y = 28
y=4
When y = 4, x = 5.
………………………………………………………………………………
Simultaneous Non-Linear Equations
6.
A non-linear equation is not of the form ax + by = p.
7.
Methods of solving simultaneous non-linear equations:
•
Substitution
•
Graphical method (covered in ‘O’ level Mathematics)
Simultaneous Equations, Polynomials and Partial Fractions
3
8.
The method most commonly used to solve simultaneous non-linear equations is
• Substitution
9.
The substitution method:
Step 1: Use the linear equation to express one of the variables in terms of the other.
Step 2: Substitute it into the non-linear equation.
Step 3: Substitute the value(s) obtained in Step 2 into the linear equation to obtain
the value of the other variable.
Example 3
Solve the following pair of simultaneous equations.
3y = x + 3
y2 = 13 + 2x
Solution
3y = x + 3
––– (1)
2
y = 13 + 2x ––– (2)
From (1):
y=
1
x+1
3
––– (3)
(Use the linear equation to express y in terms of x.)
Substitute (3) into (2):
2
⎛1
⎞
⎜ x + 1⎟ = 13 + 2x
3
⎝
⎠
1 2 2
x + x + 1 = 13 + 2x
9
3
1 2 4
x – x – 12 = 0
9
3
x2 – 12x – 108 = 0
(x – 18)(x + 6) = 0
x = 18 or x = –6
When x = 18, y = 7. (Substitute the values of x into the linear equation to obtain
When x = –6, y = –1. the corresponding values of y.)
∴ x = 18, y = 7
4
UNIT 1
or x = –6, y = –1
Example 4
Solve the simultaneous equations
x2 – 2y2 = –17,
x – y = – 4.
Solution
x2 – 2y2 = –17 ––– (1)
x – y = – 4 ––– (2)
From (2),
y = x + 4 ––– (3) (Use the linear equation to express y in terms of x.)
Substitute (3) into (1):
x2 – 2(x + 4)2 = –17 (Substitute the linear equation into the non-linear
2
x – 2x2 – 16x – 32 = –17 equation.)
–x2 – 16x – 15 = 0
x2 + 16x + 15 = 0
(x + 1)(x + 15) = 0 (Factorise the quadratic expression.)
x = –1 or x = –15
When x = –1, y = 3.
When x = –15, y = –11.
(Substitute the values of x into the linear equation
to obtain the corresponding values of y.)
 x = −1, y = 3 or x = −15, y = −11
Simultaneous Equations, Polynomials and Partial Fractions
5
Example 5
The line 2x + y = 5 meets the curve x2 + y2 + x + 12y – 29 = 0 at the points A and B.
Find the coordinates of A and B.
Solution
2x + y = 5 — (1)
x2 + y2 + x + 12y – 29 = 0 — (2)
From (1),
y = 5 – 2x — (3) (Use the linear equation to express y in terms of x.)
Substitute (3) into (2):
x2 + (5 – 2x)2 + x + 12(5 – 2x) – 29 = 0 (Substitute the linear equation
2
x + 25 – 20x + 4x2 + x + 60 – 24x – 29 = 0 into the non-linear equation.)
5x2 – 43x + 56 = 0
(5x – 8)(x – 7) = 0 (Factorise the quadratic expression.)
3
x = 1 or x = 7
5
4
3
When x = 1 , y = 1 . (Substitute the values of x into the linear equation
5
5
to obtain the corresponding values of y.)
When x = 7, y = –9.
⎛ 3 4 ⎞
∴ The coordinates of A and B are ⎜1 , 1 ⎟ and (7, –9).
⎝ 5 5 ⎠
6
UNIT 1
Definitions
10. A polynomial in x is a mathematical expression of a sum of terms, each of the form
axn, where a is a constant and n is a non-negative integer. It is usually denoted as f(x).
i.e. f(x) = anxn + an – 1 xn – 1 + an – 2 xn – 2 + … + a2 x2 + a1x + a0
1
11. Examples of polynomials include x3 + 2x − 1, 6x4 – x2 and −0.2x + x2 + 5x3.
2
2
1
Examples of non-polynomials include 2x2 + , 4 – x and x + x 3.
x
12. an, an – 1, …, a0 are coefficients.
a0 is also called the constant term.
13. The degree (or order) of a polynomial in x is given by the highest power of x.
For example, the degree of 6x3 − 2x2 + x − 8 is 3 and the degree of 1 − x + 5x4 is 4.
14. The value of f(x) at x = c is f(c).
For example, if f(x) = 2x3 + x2 − x − 4, then the value of f(x) at x = 1 is
f(1) = 2(1)3 + 12 − 1 − 4 = −2.
Identities
15. An identity is an equation in which the expression on the LHS (left-hand side) is
equal to the expression on the RHS (right-hand side).
16. Methods of finding the unknown constants in an identity:
• By substitution of special values of x
• By comparing coefficients
Simultaneous Equations, Polynomials and Partial Fractions
7
Example 6
It is given that for all values of x, 2x3 + 5x2 – x – 2 = (Ax + 3)(x + B)(x – 1) + C.
Find the values of A, B and C.
Solution
Let x = 1: 2(1)3 + 5(1)2 – 1 – 2 = (A + 3)(1 + B)(1 – 1) + C
C=4
Let x = 0: 2(0)3 + 5(0)2 – 0 – 2 = (0 + 3)(0 + B)(0 – 1) + 4
B=2
Comparing coefficients of x3,
A=2
 A = 2, B = 2 and C = 4
(Letting x be 1 leaves
us with 1 unknown, C.)
(Letting x be 0 leaves
us with 1 unknown, B.)
Example 7
Given that 2x3 + 3x2 – 14x – 5 = (2x – 3)(x + 3)Q(x) + ax + b, where Q(x) is a
polynomial, find the value of a and of b.
Solution
Let x = –3: 2(–3)3 + 3(–3)2 – 14(–3) – 5 = –3a + b
10 = –3a + b
3a – b = –10 — (1)
3
2
⎛ 3⎞
⎛ 3⎞
3 ⎛ 3⎞
3
: 2 ⎜ ⎟ + 3 ⎜ ⎟ – 14 ⎜ ⎟ – 5 = a + b
2
2
2
2
2
⎝ ⎠
⎝ ⎠
⎝ ⎠
25 3
–
= a+b
2 2
3a + 2b = –25 — (2)
(2) − (1): 3b = −15
b = −5
a = −5
 a = –5, b = –5
Let x =
8
UNIT 1
Long Division
17. When 3x3 + 4x2 – 6x + 3 is divided by x – 1,
• the dividend is 3x3 + 4x2 – 6x + 3
• the quotient is 3x2 + 7x + 1
• the divisor is x – 1
• the remainder is 4.
Divisor
3x 2 + 7x + 1
x – 1 3x 3 + 4 x 2 – 6x + 3
Quotient
Dividend
–(3x 3 – 3x 2 )
7x 2 – 6x + 3
–(7x 2 – 7x)
x+3
–(x – 1)
4
Remainder
18. Dividend = Quotient × Divisor + Remainder
19. The order of the remainder is always at least one degree less than that of the divisor.
20. The process of long division is stopped when the degree of the remainder is less
than the degree of the divisor.
Synthetic Method
21. The synthetic method can be used to divide a polynomial by a linear divisor.
To divide 3x3 + 4x2 – 6x + 3 by x – 1,
1 3 4 –6
3 7
3
7 1
3 Coefficients of
the Dividend
1
4
Coefficients of
the Quotient
Remainder
Simultaneous Equations, Polynomials and Partial Fractions
9
Remainder Theorem
22. The Remainder Theorem states that when a polynomial f(x) is divided by ax – b,
⎛b ⎞
the remainder is f ⎜ ⎟ .
⎝a ⎠
23. If f(x) is divided by a quadratic divisor, then the remainder is a linear function or a
constant.
Example 8
Find the remainder when x3 – 2x2 + 3x – 1 is divided by x – 1.
Solution
Let f(x) = x3 – 2x2 + 3x – 1.
By Remainder Theorem,
The remainder is f(1) = (1)3 – 2(1)2 + 3(1) – 1
=1–2+3–1
=1
10
UNIT 1
Example 9
Given that f(x) = ax3 – 8x2 – 9x + b is exactly divisible by 3x – 2 and leaves a
remainder of 6 when divided by x, find the value of a and of b.
Solution
⎛2 ⎞
Since f ⎜ ⎟ = 0,
⎝ 3⎠
3
2
⎛2 ⎞
⎛2 ⎞
⎛2 ⎞
a ⎜ ⎟ – 8 ⎜ ⎟ – 9 ⎜ ⎟ + b = 0
⎝ 3⎠
⎝ 3⎠
⎝ 3⎠
8 a – 32 – 6 + b = 0
9
27
8a – 96 – 162 + 27b = 0
8a + 27b = 258 — (1)
Since f(0) = 6,
a(0)3 – 8(0)2 – 9(0) + b = 6
b = 6 — (2)
Substitute b = 6 into (1):
8a + 27(6) = 258
a = 12
 a = 12, b = 6
Simultaneous Equations, Polynomials and Partial Fractions
11
Example 10
Given that f(x) = 6x3 + 7x2 – x + 3, find the remainder when f(x) is divided by x + 1.
Solution
Method 1: Long division
6x 2 + x – 2
x + 1 6x 3 + 7x 2 – x + 3
–(6x 3 + 6x 2 )
x2 – x + 3
–(x 2 + x)
– 2x + 3
–(–2x – 2)
5
 The remainder is 5.
Method 2: Synthetic method
 The remainder is 5.
–1 6
7
–6
6 1
–1
–1
–2
3
2
5
Method 3: Remainder Theorem
f(x) = 6x3 + 7x2 – x + 3
f(–1) = 6(–1)3 + 7(–1)2 – (–1) + 3
=5
 The remainder is 5.
………………………………………………………………………………
Factor Theorem
24. The Factor Theorem states that when a polynomial f(x) is divided by ax – b and
⎛b ⎞
that f ⎜ ⎟ = 0, then ax – b is a factor of f(x).
⎝a ⎠
⎛b ⎞
25. Conversely, if ax – b is a factor of f(x), then f ⎜ ⎟ = 0 and f(x) is divisible by ax – b.
⎝a ⎠
12
UNIT 1
Example 11
Given that x + 2 is a factor of x3 + ax2 – x + 4, calculate the value of a.
Solution
Let f(x) = x3 + ax2 – x + 4.
Since x + 2 is a factor of f(x), by Factor Theorem,
f(–2) = 0
(–2) + a(–2) – (–2) + 4 = 0
–8 + 4a + 2 + 4 = 0
–2 + 4a = 0
1
a=
2
3
2
Example 12
Prove that x + 2 is a factor of 4x3 – 13x + 6. Hence solve the equation 4x3 – 13x + 6 = 0.
Solution
Let f(x) = 4x3 – 13x + 6.
(To prove that x + 2 is a factor of f(x),
3
f(–2) = 4(–2) – 13(–2) + 6 we need to show that f(–2) = 0.)
=0
 x + 2 is a factor of 4x3 – 13x + 6.
Now f(x) = 4x3 – 13x + 6 = (x + 2)(4x2 + kx + 3), where k is a constant.
Comparing coefficients of x2,
0=8+k
k = –8
i.e. f(x) = (x + 2)(4x2 – 8x + 3)
= (x + 2)(2x – 1)(2x – 3)
To solve 4x3 – 13x + 6 = 0,
(x + 2)(2x – 1)(2x – 3) = 0
1
3
 x = –2 or x =
or x =
2
2
Simultaneous Equations, Polynomials and Partial Fractions
13
Example 13
Given that 4x3 + ax2 + bx + 2 is exactly divisible by x2 – 3x + 2, find the value of a
and of b. Hence sketch the graph of y = 4x3 + ax 2 + bx + 2 for the values of a and
b found.
Solution
Let f(x) = 4x3 + ax2 + bx + 2.
Since x2 – 3x + 2 = (x – 1)(x – 2),
f(x) is exactly divisible by (x – 1)(x – 2), (Factorise the quadratic divisor.)
i.e. f(1) = 0 and f(2) = 0.
When f(1) = 0,
4(1)3 + a(1)2 + b(1) + 2 = 0
4+a+b+2=0
a + b = –6 — (1)
When f(2) = 0,
4(2)3 + a(2)2 + b(2) + 2 = 0
32 + 4a + 2b + 2 = 0
4a + 2b = –34
2a + b = –17 — (2)
(2) − (1):
a = –11
b = 5
 a = –11, b = 5
f(x) = 4x3 – 11x2 + 5x + 2 = (x2 – 3x + 2)(px + q)
Comparing coefficients of x3,
p = 4
Comparing constants,
2 = 2q
q = 1
f(x) = (x2 – 3x + 2)(4x + 1)
= (x – 1)(x – 2)(4x + 1)
When f(x) = 0,
x = 1 or x = 2 or x = –
14
UNIT 1
1
. (It is a good practice to find the intercepts with
4
the coordinate axes before sketching the graph.)
When x = 0,
f(0) = 2.
y
(0, 2)
⎛ 1 ⎞
⎜– , 0⎟
⎝ 4 ⎠
O
y = f(x)
(1, 0)
(2, 0)
x
………………………………………………………………………………
Factorisation of Cubic Expressions
26. A cubic expression is of the form ax3 + bx2 + cx + d.
27. Cubic expressions are factorised into:
• 3 linear factors, i.e. (px + q)(rx + s)(tx + u), or
• 1 linear and 1 quadratic factor, i.e. (px + q)(rx2 + sx + t), where rx2 + sx + t cannot
be factorised into 2 linear factors
28. Methods of factorising cubic expressions:
• Trial and error
• Long division
• Synthetic method
• Comparing coefficients
29. Sum and difference of cubes:
• Sum of cubes: a3 + b3 = (a + b)(a2 – ab + b2)
• Difference of cubes: a3 – b3 = (a – b)(a2 + ab + b2)
Solving Cubic Equations
30. To solve the equation f(x) = 0,
Step 1: Factorise f(x) using the Factor Theorem.
Step 2: Use the synthetic method or compare coefficients to factorise f(x)
completely.
Step 3: Equate each factor to zero and use general solution where necessary.
Simultaneous Equations, Polynomials and Partial Fractions
15
Example 14
Solve the equation 2x3 + x2 – 5x + 2 = 0.
Solution
Let f(x) = 2x3 + x2 – 5x + 2.
f(1) = 2 + 1 – 5 + 2
=0
∴ (x – 1) is a factor of f(x).
By long division,
2x 2 + 3x – 2
x – 1 2x 3 + x 2 – 5x + 2
– (2x 3 – 2x 2 )
3x 2 – 5x + 2
–(3x 2 – 3x)
–2x + 2
–(–2x + 2)
0
f(x) = (x – 1)(2x2 + 3x – 2)
= (x – 1)(2x – 1)(x + 2)
When f(x) = 0,
1
x = 1 or x =
or x = –2.
2
16
UNIT 1
Example 15
In the cubic polynomial f(x), the coefficient of x3 is 4 and the roots of f(x) = 0
1
are 3, and – 4.
2
(i) Express f(x) as a cubic polynomial in x with integer coefficients.
(ii) Find the remainder when f(x) is divided by 2x – 5.
(iii) Solve the equation f x = 0.
( )
Solution
1
(i) Since the roots of f(x) = 0 are 3, and – 4, the factors of f(x) are x – 3,
2
2x – 1 and x + 4.
Given also that the coefficient of x3 is 4, f(x) = 2(x – 3)(2x – 1)(x + 4)
= 4x3 + 2x2 – 50x + 24
3
2
⎛5 ⎞
⎛5 ⎞
⎛5 ⎞
⎛5 ⎞
(ii) f ⎜ ⎟ = 4 ⎜ ⎟ + 2 ⎜ ⎟ – 50 ⎜ ⎟ + 24
⎝2 ⎠
⎝2 ⎠
⎝2 ⎠
⎝2 ⎠
= –26
∴ The remainder is –26.
(iii) Since f(x) = 2(x – 3)(2x – 1)(x + 4),
f
( x) = 2 Q x – 3R Q2
When f ( x ) = 0,
x – 1R Q x + 4R (Note that x is replaced with x .)
2 Q x – 3R Q2 x – 1R Q x + 4R = 0
x – 3 = 0
or 2 x – 1 = 0
1
x =3
x=
2
1
x = 9
x=
4
 x = 9 or x =
or
x +4=0
x = –4 (no real solution)
1
4
Simultaneous Equations, Polynomials and Partial Fractions
17
Algebraic Fractions
31. An algebraic fraction is the ratio of two polynomials of the form
and D(x) are polynomials in x.
Proper and Improper Fractions
32. If the degree of P(x) is less than the degree of D(x),
P(x)
, where P(x)
D(x)
P(x)
is a proper fraction.
D(x)
P(x)
33. If the degree of P(x) is more than or equal to the degree of D(x),
is an
D(x)
improper fraction.
P(x)
, we can make use of long division to
D(x)
P(x)
R(x)
R(x)
obtain
= Q(x) +
, where Q(x) is a polynomial and
is a proper
D(x)
D(x)
D(x)
algebraic fraction.
34. From an improper algebraic fraction
35. To express a compound algebraic fraction into partial fractions:
Step 1: Determine if the compound fraction is proper or improper. If it is
improper, perform long division (or use the synthetic method if the
denominator is linear).
Step 2: Ensure that the denominator is completely factorised.
Step 3: Express the proper fraction in partial fractions according to the cases below.
Step 4: Solve for unknown constants by substituting values of x and/or comparing
coefficients of like terms and/or using the “Cover-Up Rule”.
Rules of Partial Fractions
36.
18
Case
Denominator
of fraction
Algebraic
fraction
Expression used
1
Linear factors
mx + n
(ax + b)(cx + d)
A
B
+
ax + b cx + d
2
Repeated linear
factors
mx + n
(ax + b)(cx + d)2
A
B
C
+
+
ax + b cx + d (cx + d)2
3
Quadratic factor
which cannot be
factorised
mx + n
(ax + b)(x 2 + c 2 )
Bx + C
A
+
ax + b x 2 + c 2
UNIT 1
Example 16
Express
7 – 2x
in partial fractions.
x +x–6
2
Solution
First factorise the denominator to get the algebraic fraction in the form
mx + n
of
.
(ax + b)(cx + d)
x2 + x – 6 = (x + 3)(x – 2)
7 – 2x
7 – 2x
=
(x + 3)(x – 2)
x2 + x – 6
Then, let
7 – 2x
A
B
=
+
.
(x + 3)(x – 2)
x+3
x–2
Multiply throughout by (x + 3)(x – 2),
7 – 2x = A(x – 2) + B(x + 3)
Let x = 2: 7 – 2(2) = 5B (Substituting x = 2 leaves us with 1 unknown, B.)
3
B=
5
Let x = –3: 7 – 2(–3) = A(–5)
13
A=–
5
∴
(Substituting x = –3 leaves us with 1 unknown, A.)
7 – 2x
13
3
= –
+
5(x + 3)
5(x – 2)
x2 + x – 6
Simultaneous Equations, Polynomials and Partial Fractions
19
Example 17
Express
x4 + 9
in partial fractions.
x 3 + 3x
Solution
First we need to perform long division on
x
x 3 + 3x x 4 + 0x 2 + 9
–(x 4 + 3x 2 )
–3x 2 + 9
–3x 2 + 9
x4 + 9
=
x
+
x 3 + 3x
x 3 + 3x
–3x 2 + 9
=x+
x(x 2 + 3)
–3x 2 + 9
Bx + C
A
=
+ 2
.
2
x
x(x + 3)
x +3
Multiply throughout by x(x2 + 3),
–3x2 + 9 = A(x2 + 3) + (Bx + c)x
Let x = 0 : 9 = 3A
A=3
Comparing coefficients of x2,
–3 = A + B
=3+B
B = –6
Comparing coefficients of x,
C=0
Let
∴
20
x4 + 9
6x
3
=x+
– 2
x
x +3
x 3 + 3x
UNIT 1
x4 + 9
.
x 3 + 3x
Example 18
Express
2x 3 – 2x 2 – 24 x – 7
in partial fractions.
x 2 – x – 12
Solution
By long division,
2x
2
3
x
–
x
–
12
2x
– 2x 2 – 24 x – 7
–(2x 3 – 2x 2 – 24 x)
–7
3
2
2x – 2x – 24 x – 7
–7
= 2x + 2
2
x – x – 12
x – x – 12
–7
= 2x +
(x – 4)(x + 3)
–7
A
B
=
+
.
(x – 4)(x + 3)
x–4
x+3
Let
(Ignore the 2x when expressing
–7
into its partial fractions.)
(x – 4)(x + 3)
Multiply throughout by (x – 4)(x + 3),
–7 = A(x + 3) + B(x – 4)
Let x = 4: –7 = 7A
A = –1
Let x = –3: –7 = –7B
B=1

1
1
2x 3 – 2x 2 – 24 x – 7
= 2x –
+
2
x
–
4
x
+
3
x – x – 12
Simultaneous Equations, Polynomials and Partial Fractions
21
Example 19
Express
8x 2 – 5x + 2
in partial fractions.
(3x + 2)(x 2 + 4)
Solution
Let
Bx + C
A
8x 2 – 5x + 2
=
+ 2
.
2
3x + 2
x +4
(3x + 2)(x + 4)
Multiply throughout by (3x + 2)(x2 + 4),
8x2 – 5x + 2 = A(x2 + 4) + (Bx + C)(3x + 2)
2 80
40
:
=
A
3 9
9
A=2
Let x = 0: 2 = 4A + 2C
= 4(2) + 2C
2C = – 6
C = –3
Comparing coefficients of x2,
8 = A + 3B
= 2 + 3B
3B = 6
B=2
Let x = –

22
8x 2 – 5x + 2
2x – 3
2
=
+ 2
2
3x + 2
(3x + 2)(x + 4)
x +4
UNIT 1
(Note
that x2 + 4 cannot be factorised
into 2 linear factors.)
Example 20
Express
9 – 4x
in partial fractions.
(2x + 3)(x – 1)2
Solution
Let
9 – 4x
A
B
C
=
+
+
.
2
2x
+
3
x
–
1
(x
–
1)2
(2x + 3)(x – 1)
Multiply throughout by (2x + 3)(x – 1)2,
9 – 4x = A(x – 1)2 + B(x – 1)(2x + 3) + C(2x + 3)
Let x = 1: 5 = 5C
C=1
3
25
Let x = – : 15 =
A
2
4
12
A=
5
Comparing coefficients of x2,
0 = A + 2B
12
0=
+ 2B
5
12
2B = –
5
6
B=–
5

9 – 4x
(2x + 3)(x – 1)2
=
12
5(2x + 3)
–
6
5(x – 1)
+
1
(x – 1)2
………………………………………………………………………………
37. Cover-Up Rule
The “Cover-Up Rule” is a method to find the unknown numerators of partial
fractions.
Given that
P(x)
A
B
=
+
, where P(x) is a linear polynomial,
(ax + b)(cx + d)
ax + b
cx + d
⎛ b ⎞
⎛ d ⎞
P ⎜– ⎟
P ⎜– ⎟
a
⎝
⎠
⎝ c ⎠ .
A=
and B =
⎛ b ⎞
⎛ d ⎞
c ⎜– ⎟ + d
a ⎜– ⎟ + b
⎝ a ⎠
⎝ c ⎠
Simultaneous Equations, Polynomials and Partial Fractions
23
Example 21
Express
3x – 1
in partial fractions.
(x + 3)(x – 2)
Solution
Let
3x – 1
A
B
=
+
.
(x + 3)(x – 2)
x+3
x–2
Method 1: Substitution
3x – 1
A
B
=
+
(x + 3)(x – 2)
x+3
x–2
Multiply throughout by (x + 3)(x – 2),
3x – 1 = A(x – 2) + B(x + 3)
Let x = 2: 5 = 5B (Letting x be 2 leaves us with 1 unknown, B.)
B=1
Let x = –3: –10 = –5A (Letting x be –3 leaves us with 1 unknown, A.)
A=2
3x – 1
2
1

=
+
(x + 3)(x – 2)
x+3
x–2
Method 2: Cover-Up Rule
Using the Cover-Up Rule,
3(–3) – 1
3(2) – 1
and B =
–3 – 2
2+3
= 2 =1
3x – 1
2
1

=
+
(x + 3)(x – 2)
x+3
x–2
A=
24
UNIT 1
UNIT
2
Quadratic Equations,
Inequalities and
Modulus Functions
Relationships between the Roots and Coefficients
of a Quadratic Equation
1.
If α and β are the roots of the quadratic equation ax2 + bx + c = 0,
b
a
c
Product of roots, αβ =
a
Sum of roots, α + β = –
i.e. x2 – (α + β )x + αβ = 0
2.
In general,
x2 – (sum of roots)x + (product of roots) = 0
3.
Some useful identities
(i) α2 + β 2 = (α + β)2 – 2αβ
(ii) (α – β)2 = (α + β )2 – 4αβ
(iii) α4 – β 4 = (α2 + β 2)(α + β )(α – β )
(iv) α4 + β 4 = (α2 + β2) 2 – 2α2β 2
(v) α3 – β3 = (α – β )[(α + β )2 – αβ]
(vi) α3 + β 3 = (α + β)[(α + β )2 – 3αβ ]
Quadratic Equations, Inequalities and Modulus Functions
25
Example 1
The roots of the quadratic equation 2x2 – 5x = 4 are α and β.
Find
β
α
+
(i) α2 + β 2, (ii)
.
2 β 2a
Solution
From 2x2 – 5x = 4, we have 2x2 – 5x – 4 = 0.
α+β=–
5
–5
=
2
2
–4
= –2
2
α2 + β2
β
α
+
(i) α2 + β2 = (α + β)2 – 2αβ
(ii)
=
2 β 2a
2aβ
2
⎛5 ⎞
41
= ⎜ ⎟ – 2(–2) =
÷ 2(–2)
4
⎝2 ⎠
αβ =
=
41
4
=
41
16
Example 2
Using your answers in Example 1, form a quadratic equation with integer α α β+ β
coefficients whose roots are
+and
.
2 β2 β 2a 2a
Solution
β
α
41
+
=–
2 β 2α
16
β
α
1
×
=
Product of new roots,
2 β 2α 4
⎛ 41⎞
1
∴ New equation is x2 – ⎜– ⎟ x + = 0
16
4
⎝
⎠
i.e. 16x2 + 41x + 4 = 0.
Sum of new roots,
26
UNIT 2
Example 3
If α and β are the roots of the equation 2x2 + 5x – 12 = 0, where α . β,
find the value of each of the following.
1 1
+ (i)
(ii) α2 + β2
α β
Solution
b
a
5
= – 2
1
= –2 2
α+β=–
(i)
αβ =
c
a
= –
12
2
= –6
α+β
1 1
+
=
α β
αβ
1
2
=
–6
–2
=
5
12
(ii) α2 + β2 = (α + β)2 – 2αβ
2
⎛ 1⎞
= ⎜–2 ⎟ – 2(–6)
⎝ 2⎠
= 18
1
4
Quadratic Equations, Inequalities and Modulus Functions
27
Maximum and Minimum Values of Quadratic
Functions
4.
The quadratic function ax2 + bx + c can be expressed as a(x + h)2 + k.
a>0
y
a<0
y = a(x + h)2 + k
y
(–h, k)
y = a(x + h)2 + k
(–h, k)
Minimum value = k, when x = –h
Minimum point: (–h, k)
x
x
Maximum value = k, when x = –h
Maximum point: (–h, k)
Sketching of Quadratic Graphs
5.
Method of sketching a quadratic graph:
Step 1: Determine the shape of the graph from a.
Step 2: Express the function as a(x + h)2 + k to get the coordinates of the maximum or
minimum point.
Step 3: Substitute x = 0 to find the y-intercept.
Step 4: Substitute y = 0 to find the x-intercept(s), if the roots are real.
28
UNIT 2
Example 4
Sketch the function y = x2 – 1.
Solution
Step 1: Since y = x2 – 1 is a quadratic function and a is positive, the graph is
U-shaped.
Step 2: Comparing with the form a(x + h)2 + k, we get a = 1, which is greater than 0
so it has a minimum point.
From the function y = x2 – 1, h = 0, k = –1. (We can express x2 – 1
∴ Minimum point = (0, –1) as (x + 0)2 + (–1) and compare
with the form a(x + h)2 + k.)
Step 3: When x = 0, f(x) = –1.
y
Step 4: When y = 0, x = 1 and –1.
f(x) = x2 – 1
–1
0
1
x
–1
Quadratic Equations, Inequalities and Modulus Functions
29
Quadratic Inequalities
6.
If (x – a)(x – b) . 0, If (x – a)(x – b) > 0,
then x , a or x . b. then x  a or x  b.
(x –– a)(x
a)(x –– b)
b) >> 00
(x
xx << aa
7.
aa
bb
(x –– a)(x
a)(x –– b)
b) 00
(x
xx >> bb
bb
xx aa
bb
xx
If (x – a)(x – b) , 0,
If (x – a)(x – b)  0,
then a , x , b. then a  x  b.
a < x <a b< x < b
a
b
a
b
(x –– a)(x
(x – a)(x
b) <–0b) < 0
30
aa
xx xx
UNIT 2
x
x
a x ab x b
a
b
a
b
x
x
Example 5
Find the range of values of x for which 3x2 – 4x + 6  7x.
Solution
3x2 – 4x + 6  7x (When solving quadratic inequalities, ensure that the RHS of
3x2 – 11x + 6  0 the inequality is zero before factorising the expression on the
(3x – 2)(x – 3)  0 LHS.)
2
3
∴ Range of values of x is
3
x
2
x3
3
………………………………………………………………………………
Roots of a Quadratic Equation
8.
The roots of a quadratic equation ax2 + bx + c = 0 are given by x =
9.
b2 – 4ac is called the discriminant.
–b ± b 2 – 4ac
.
2a
10. A quadratic equation has no real roots when b2 – 4ac , 0.
Given a quadratic expression ax2 – bx + c,
it is found that:
given that b2 – 4ac , 0 and a . 0, ax2 – bx + c . 0 for all real values of x, and
given that b2 – 4ac , 0 and a , 0, ax2 – bx + c , 0 for all real values of x.
Quadratic Equations, Inequalities and Modulus Functions
31
Example 6
Is the quadratic expression 5x2 + 4x + 1 greater than zero for all real values of x?
Solution
Discriminant = 42 – 4(5)(1)
= –4
Since b2 – 4ac , 0 and a . 0, 5x2 + 4x + 1 . 0 for all real values of x.
Example 7
Find the range of values of k for which the equation 2x2 + 5x – k = 0 has no real
roots.
Solution
2x2 + 5x – k = 0
a = 2, b = 5, c = –k
For the equation to have no real roots,
b2 – 4ac , 0
2
5 – 4(2)(–k) , 0
8k , –25
25
k,–
8
32
UNIT 2
Conditions for the Intersection of a Line and a
Quadratic Curve
11.
b2 – 4ac
>0
Nature
of
roots
Intersection of
y = ax2 + bx + c
with the x-axis
Intersection of
quadratic curve with a
straight line
2 real and
distinct
roots
y = ax2 + bx + c cuts the
x-axis at 2 distinct points
Line intersects the curve
at two distinct points
y
x
a>0
y
x
a<0
=0
2 real and
equal
roots
y = ax2 + bx + c touches
the x-axis
y
y
a>0
Line is a tangent to the
curve
x
x
a<0
Quadratic Equations, Inequalities and Modulus Functions
33
<0
No real
roots
y = ax2 + bx + c lies
entirely above or entirely
below the x-axis i.e.
curve is always positive
(a . 0) or always negative
(a , 0)
y
y
a>0
x
x
a<0
34
UNIT 2
Line does not intersect
the curve
Example 8
Find the range of values of k given that the straight line y = x – k cuts the curve
y = kx2 + 9x at two distinct points.
Solution
y=x–k
––– (1)
y = kx2 + 9x ––– (2)
Substitute (1) into (2): (Substitute (1) into (2) to obtain a quadratic equation in x.)
x – k = kx2 + 9x
2
kx + 8x + k = 0
Since the straight line cuts the curve at two distinct points,
Discriminant . 0
82 – 4(k)(k) . 0
64 – 4k2 . 0
k2 – 16 , 0
(k + 4)(k – 4) , 0
(Remember to invert the inequality sign when dividing
by a negative number.)
–4
4
k
∴ Range of values of k is –4 , k , 4
Quadratic Equations, Inequalities and Modulus Functions
35
Example 9
Find the range of values of m for which the line y = 5 – mx does not intersect the
curve x2 + y2 = 16.
Solution
y = 5 – mx ––– (1)
x2 + y2 = 16 ––– (2)
Substitute (1) into (2):
x2 + (5 – mx)2 = 16
2
2 2
x + m x – 10mx + 25 = 16
(1 + m2)x2 – 10mx + 9 = 0
Since the line does not intersect the curve,
Discriminant , 0
2
(–10m) – 4(1 + m2)(9) , 0
100m2 – 36 – 36m2 , 0
64m2 – 36 , 0
16m2 – 9 , 0
(4m + 3)(4m – 3) , 0
3
–
4
∴ Range of values of m is –
36
UNIT 2
3
4
3
3
,m,
4
4
m
Absolute Valued Functions
| |
12. The absolute value of a function f(x), i.e. f(x) , refers to the numerical value of f(x).
| | –f(x) if f(x) , 0
|f(x)|  0 for all values of x.
13. f(x) =
14.
f(x) if f(x)  0
Example 10
|
|
Solve 4x – 3 = 2x.
Solution
|4x4x––3­3| == 2x2x
or 4x – 3 = –2x
2x = 3
6x = 3
3
1
x=
x=
2
2
3
1
∴x=
or x =
2
2
Example 11
|
|
Solve 2x – 3 = 15.
Solution
|2x2x––3­3| == 1515
or
2x = 18
x=9
∴ x = 9 or x = –6
2x – 3 = –15
2x = –12
x = –6
Quadratic Equations, Inequalities and Modulus Functions
37
Example 12
|
| | |
Solve 2x – 5 = 4 – x .
Solution
|2x – 5| = |4 – x­ |
2x – 5 = 4 – x
or
3x = 9
x=3
∴ x = 3 or x = 1
2x – 5 = –(4 – x)
= –4 + x
x=1
Example 13
|
|
Solve x2 – 3 = 2x.
Solution
|x – 3| = 2x
2
x2 – 3 = 2x
x – 2x – 3 = 0
(x – 3)(x + 1) = 0
x = 3 or x = –1
2
or
x2 – 3 = –2x
x + 2x – 3 = 0
(x + 3)(x – 1) = 0
x = –3 or x = 1
2
Checking the solutions, x = 3 or x = 1. (Substitute your answers into the original
equation to check for any extraneous
solutions.)
38
UNIT 2
Example 14
|
|
Solve 2x2 – 5x = x.
Solution
|2x – 5x| = x
2
2x2 – 5x = x
or
2
2x – 6x = 0
2x(x – 3) = 0
x = 0 or x = 3
2x2 – 5x = –x
2x2 – 4x = 0
2x(x – 2) = 0
x = 0 or x = 2
∴ x = 0, x = 2 or x = 3
………………………………………………………………………………
| |
Graphs of y = f(x)
| |
15. Method of sketching the graph of y = f(x) :
Step 1: Sketch the graph of y = f(x).
Step 2: The part of the graph below the x-axis is reflected in the x-axis.
Quadratic Equations, Inequalities and Modulus Functions
39
Example 15
Sketch the graph of y = 2x.
Hence, sketch the graph of y = 2x .
Solution
Sketch the graph y = 2x.
y
y = 2x
2
1
–2
–1
O
1
2
x
–1
–2
To draw y = 2x , reflect the part of the graph that lies below the x-axis.
y
y = 2x
2
1
–2
–1
O
1
–1
–2
40
UNIT 2
2
x
Example 16
|
|
Sketch the graph of y = 2x + 1 for the domain –1 < x < 1 and state the
corresponding range.
Solution
y
3
|
(It is necessary to find the coordinates of the
critical points.)
|
y = 2x + 1
1
–1 – 1 O
2
1
x
∴ Range is 0 < y < 3
Quadratic Equations, Inequalities and Modulus Functions
41
Example 17
|
|
Sketch the graph of y = x2 – 2x – 3 for –2 < x < 3. State the corresponding range.
Solution
y
5
4
3
2
–2
|
1
–1
|
y = x2 – 2x – 3
O
1
3
x
(Note that x2 – 2x – 3
= (x – 3)(x + 1).)
∴ Range is 0 < y < 5
………………………………………………………………………………
|
|
16. If a function is defined as y = a bx + c + d,
(a) If a . 0, it is a V-shaped graph.
If a , 0, it is an inverted V-shaped graph.
|
|
|
|
(b) If d . 0, y = a bx + c is translated up by d units.
||
If d , 0, y = a bx + c is translated down by d units.
42
UNIT 2
Example 18
Sketch the graph of y = 2 x – 1.
Solution
Step 1: Sketch the graph y = 2 x .
y
y=2x
2
1
–2
–1
O
–1
1
2
x
–2
Step 2: Translate the graph down by 1 unit.
y
2
y = 2x – 1
1
–2
–1
O
–1
1
2
x
–2
Quadratic Equations, Inequalities and Modulus Functions
43
UNIT
Binomial Theorem
3
Binomial Theorem
1.
For a positive integer n,
⎛ n ⎞ n – 1 ⎛ n ⎞ n – 2 2
⎛ n
n
(a + b)n = a + ⎜⎜ ⎟⎟ a b + ⎜⎜ ⎟⎟ a b + … + ⎜⎜
⎝ 1 ⎠
⎝ 2 ⎠
⎝ r
⎛ n ⎞
n(n – 1)…n(n – r + 1)
n!
where ⎜⎜ ⎟⎟ =
=
r!
r!(n
–
r)!
⎝ r ⎠
2.
Number of terms in the expansion of (a + b)n is n + 1
3.
Special case:
When a = 1,
⎛
(1 + b)n = 1 + ⎜ n
⎝ 1
44
⎞ n – r r
⎟⎟ a b + … + b n
⎠
UNIT 3
⎛ n⎞ r
⎞
⎛ n⎞ 2
n
⎟b + ⎜ 2 ⎟b +…+ ⎜ r ⎟b +…+b
⎝ ⎠
⎠
⎝ ⎠
Example 1
Find the value of k and of n given that (1 + kx)n = 1 + 48x + 1008x2 + ... .
Solution
⎛ n ⎞
⎛ n ⎞
(1 + kx)n = 1 + ⎜⎜ ⎟⎟ (kx) + ⎜⎜ ⎟⎟ (kx)2 + … (Use the expansion of (1 + b)n.)
⎝ 1 ⎠
⎝ 2 ⎠
n(n – 1) 2 2
= 1 + nkx +
kx +…
2
By comparing coefficients,
(Compare coefficients to obtain a pair of simultaneous
x: nk = 48
48
k=
— (1) equations.)
n
n(n – 1)k 2
x2:
= 1008
2
n(n – 1)k2 = 2016
— (2)
Substitute (1) into (2):
2304
n(n – 1) 2 = 2016
n
2304n – 2304 = 2016n
n=8
k=6
∴ k = 6, n = 8
Binomial Theorem
45
Example 2
Write down the first 4 terms in the expansion of (1 + 2x)7 in ascending powers of x.
Hence, find the coefficient of x3 in the expansion of (1 + 2x + 3x2)(1 + 2x)7.
Solution
⎛ 7
(1 + 2x)7 = 1 + ⎜⎜
⎝ 1
⎞
⎛ 7 ⎞
⎛ 7 ⎞
⎟⎟ (2x) + ⎜⎜ ⎟⎟ (2x)2 + ⎜⎜ ⎟⎟ (2x)3 + … (Use the expansion
⎠
⎝ 3 ⎠
⎝ 2 ⎠
of (1 + b)n.)
2
3
= 1 + 14x + 84x + 280x + …
(1 + 2x + 3x2)(1 + 2x)7 = (1 + 2x + 3x2)(1 + 14x + 84x2 + 280x3 + … )
= … + 280x3 + 168x3 + 42x3 + … (There is no need to obtain
= … + 490x3 + …
terms other than x3.)
∴ Coefficient of x3 is 490
………………………………………………………………………………
The notation n!
46
4.
n! = n × (n – 1) × (n – 2) × ... × 3 × 2 × 1
5.
Some useful rules:
•
⎛ n ⎞
⎜⎜ ⎟⎟ = 1
⎝ 0 ⎠
•
⎛ n ⎞
⎜⎜ ⎟⎟ = n
⎝ 1 ⎠
•
⎛⎛nn⎞⎞ n(n
n(n––1)1)
⎜⎜⎜⎜ ⎟⎟⎟⎟===
2!
2!
2
2
⎝⎝ ⎠⎠
•
⎛⎛ nn ⎞⎞ n(n
n(n––1)(n
1)(n––2)
2)
⎜⎜⎜⎜ ⎟⎟⎟⎟===
3!
3!
⎝⎝ 33⎠⎠
•
⎛ n ⎞
⎜⎜ ⎟⎟ = 1
⎝ n ⎠
UNIT 3
Example 3
Find the value of n given that, in the expansion of (3 + 2x)n, the coefficients of x2
and x3 are in the ratio 3 : 4.
Solution
⎛ n ⎞ n – 2
⎛ n ⎞
(2x)2 + ⎜⎜ ⎟⎟ 3n – 3 (2x)3 + …
(3 + 2x)n = … + ⎜⎜ ⎟⎟ 3
⎝ 2 ⎠
⎝ 3 ⎠
⎛ n ⎞ n – 2
⎛ n ⎞
(4 x 2 )+ ⎜⎜ ⎟⎟ 3n – 3 (8x 3 ) + …
= … + ⎜⎜ ⎟⎟ 3
⎝ 2 ⎠
⎝ 3 ⎠
⎛ n ⎞ n – 2
(4)
⎜⎜ ⎟⎟ 3
3
⎝ 2 ⎠
=
4
⎛ n ⎞ n – 3
⎜⎜ ⎟⎟ 3 (8)
⎝ 3 ⎠
n(n – 1)
.3
3
2
=
4
n(n – 1)(n – 2)
.2
6
n=8
Example 4
Find the first 4 terms in the expansion of (1 + 2x)7 in ascending powers of x.
Use your result to estimate the value of 1.027.
Solution
⎛ 7 ⎞
(1 + 2x)7 = 1 + ⎜⎜ ⎟⎟ (2x) +
⎝ 1 ⎠
⎛ 7 ⎞
⎜⎜ ⎟⎟ (2x)2 +
⎝ 2 ⎠
⎛ 7 ⎞
⎜⎜ ⎟⎟ (2x)3 + …
⎝ 3 ⎠
= 1 + 14x + 84x2 + 280x3 + …
Let (1 + 2x)7 = 1.027, then x = 0.01.
1.027 = 1 + 14(0.01) + 84(0.01)2 + 280(0.01)3 + …
= 1.148 68
Binomial Theorem
47
Example 5
5
⎛ x ⎞
Expand ⎜1+ ⎟ in ascending powers of x. Hence, deduce the expansion of
⎝ 2 ⎠
5
5
5
⎛ x ⎞
⎛ x⎞ ⎛ x⎞
(i) ⎜1 – ⎟ , (ii) ⎜1+ ⎟ + ⎜1− ⎟ .
⎝ 2 ⎠
⎝ 2⎠ ⎝ 2⎠
Using your answers in (i) and (ii), find the exact value of 1.055 + 0.955.
Solution
5
⎛ 5 ⎞⎛ x ⎞ ⎛ 5 ⎞⎛ x ⎞2 ⎛ 5 ⎞⎛ x ⎞3 ⎛ 5 ⎞⎛ x ⎞4 ⎛ 5 ⎞⎛ x ⎞5
⎛ x ⎞
1
+
=
1
+
⎜⎜ ⎟⎟⎜ ⎟ + ⎜⎜ ⎟⎟⎜ ⎟ + ⎜⎜ ⎟⎟⎜ ⎟ + ⎜⎜ ⎟⎟⎜ ⎟ + ⎜⎜ ⎟⎟⎜ ⎟
⎜
⎟
⎝ 2 ⎠
⎝ 1 ⎠⎝2 ⎠ ⎝ 2 ⎠⎝2 ⎠ ⎝ 3 ⎠⎝2 ⎠ ⎝ 4 ⎠⎝2 ⎠ ⎝ 5 ⎠⎝2 ⎠
5
5
5
5 4
1 5
= 1 + x + x2 + x3 +
x +
x
2
2
4
16
32
(i)
5
5
⎛ x ⎞ ⎡ ⎛ x ⎞⎤
⎜1 – ⎟ = ⎢1 + ⎜– ⎟⎥
⎝ 2 ⎠ ⎣ ⎝ 2 ⎠⎦
5
5
5
5
1 5
= 1 – x + x2 – x3 + x4 –
x
2
2
4
16
32
⎛
(ii) ⎜1 +
⎝
5
x⎞ ⎛
⎟ + ⎜1 –
2⎠ ⎝
5
x⎞ ⎡ 5
5 2
⎟ = 1+ x + x +
2 ⎠ ⎢⎣ 2
2
⎡ 5
5
+ ⎢1 – x + x 2 –
2
2
⎣
= 2 + 5x2 +
5
5 4
x
8
5
⎛ x ⎞ ⎛ x ⎞
Let ⎜1 + ⎟ + ⎜1 – ⎟ = 1.055 + 0.955.
⎝ 2 ⎠ ⎝ 2 ⎠
By inspection,
x = 0.1
5
∴ 1.055 + 0.955 = 2 + 5(0.1)2 + (0.1)4
8
= 2.050 0625
48
UNIT 3
5 3 5 4 1 5⎤
x + x +
x
4
16
32 ⎥⎦
5 3 5 4 1 5⎤
x + x –
x
4
16
32 ⎥⎦
Example 6
Write down the expansion of (1 + p)6 in ascending powers of p. Hence, find the
first 3 terms in the expansion of (1 + 2x + 2x2)6 in ascending powers of x. Use your
result to find the value of 1.002 0026 correct to 6 decimal places.
Solution
⎛ ⎞
⎛ ⎞
⎛ ⎞
⎛ ⎞
⎛ ⎞
⎛ ⎞
(1 + p)6 = 1 + ⎜⎜ 6 ⎟⎟ p + ⎜⎜ 6 ⎟⎟ p 2 + ⎜⎜ 6 ⎟⎟ p 3 + ⎜⎜ 6 ⎟⎟ p 4 + ⎜⎜ 6 ⎟⎟ p 5 + ⎜⎜ 6 ⎟⎟ p 6
⎝ 1 ⎠
⎝ 3 ⎠
⎝ 4 ⎠
⎝ 5 ⎠
⎝ 6 ⎠
⎝ 2 ⎠
= 1 + 6p + 15p2 + 20p3 + 15p4 + 6p5 + p6
By comparing (1 + p)6 with (1 + 2x + 2x2)6,
p = 2x + 2x2
(1 + 2x + 2x2)6 = 1 + 6(2x + 2x2) + 15(2x + 2x2)2 + … (The first 3 terms consist of
= 1 + 12x + 12x2 + 60x2 + …
the constant, the term in x
2
= 1 + 12x + 72x + …
and the term in x2.)
1.002 002 = 1 + 2(0.001) + 2(0.001)2
Let x = 0.001.
∴ 1.002 0026 = 1 + 12(0.001) + 72(0.001)2 + …
= 1.012 072 (to 6 d.p.)
………………………………………………………………………………
General Term
6.
⎛ ⎞
The (r + 1)th term is ⎜⎜ n ⎟⎟ an – r br.
⎝ r ⎠
Binomial Theorem
49
Example 7
Find the 8th term in the expansion of (3 + x)12 in ascending powers of x.
Solution
⎛ 12
(r + 1)th term = ⎜⎜
⎝ r
⎞
⎟⎟ 312 – r xr
⎠
8th term = (7 + 1)th term
⎛ 12 ⎞
⎟⎟ 312 – 7 x7
= ⎜⎜
⎝ 7 ⎠
= 792 (35) x7
= 192 456x7
Example 8
In the expansion of (1 + x)n in ascending powers of x, the coefficient of the third
term is 21. Find the value of n.
Solution
⎛ n ⎞
In the expansion of (1 + x)n, the (r + 1)th term is ⎜⎜ r ⎟⎟ xr.
⎝ ⎠
⎛ n ⎞
n(n – 1)
Hence, in the expansion of (1 + x)n, the third term is ⎜⎜ ⎟⎟ x 2 =
× x2.
2
⎝ 2 ⎠
∴ The coefficient of the third term is:
n(n −1)
= 21
2
n(n – 1) = 42
n2 – n – 42 = 0
(n + 6)(n – 7) = 0
n = –6 or n = 7
∴n=7
50
UNIT 3
(Since n is a positive integer, reject n = –6.)
Term Independent of x
7.
Term independent of x refers to the constant term.
Example 9
⎛
Find the term independent of x in the expansion of ⎜x 2 +
⎝
12
1 ⎞
⎟ .
x ⎠
Solution
⎛ n ⎞ n – r r
Using Tr + 1 = ⎜⎜ ⎟⎟ a b (Recall the formula for the general term.)
⎝ r ⎠
⎛ 12 ⎞ 2 12 – r ⎛1 ⎞r
Tr + 1 = ⎜⎜
⎟⎟ (x )
⎜ ⎟
⎝ x ⎠
⎝ r ⎠
⎛ 12 ⎞ 24 – 2 r ⎛ 1 ⎞
= ⎜⎜
⎟⎟ x
⎜ r ⎟
⎝ x ⎠
⎝ r ⎠
⎛ 12 ⎞ 24 – 3r
= ⎜⎜
⎟⎟ x
⎝ r ⎠
24 – 3r = 0 (Term independent of x refers to the constant term, i.e. x0.)
r=8
⎛ 12 ⎞ 24 – 3( 8 )
∴ Term independent of x is ⎜⎜
= 495
⎟⎟ x
⎝ 8 ⎠
Example 10
Find the term independent of x in the expansion of (2x + 3)4.
Solution
⎛ 4 ⎞
In the expansion of (2x + 3)4, the (r + 1)th term is ⎜⎜ ⎟⎟ (2x)4 – r 3r.
⎝ r ⎠
For the term independent of x,
4–r=0
r=4
⎛ 4 ⎞
∴ Term independent of x is ⎜⎜ ⎟⎟ (2x)4 – 4 34 = 81
⎝ 4 ⎠
Binomial Theorem
51
UNIT
Indices, Surds and
Logarithms
4
Rules of Indices
1.
(a) am × an = am + n (c) (am)n = amn
1
(e) a–n = n a
m
(g) a n =
n
am =
n
( a)
n
m
⎛a ⎞
an
⎜ ⎟ = n , provided b ≠ 0
⎝b ⎠ b
(i)
Example 1
3
Simplify 812 × 26 ÷ 63.
Solution
3
3
812 × 26 ÷ 63 = (34 ) 2 × 26 ÷ 63
= 36 × 26 ÷ 63
= (3 × 2)6 ÷ 63
= 63
= 216
52
UNIT 4
(b) am ÷ an = am – n
(d) a0 = 1, provided a ≠ 0
(f)
1
an = n a
(h) (a × b)n = an × bn
Example 2
Simplify each of the following.
(i) 32n × 153n ÷ 5n
(ii)
25 × 5 n – 2
5n – 5n – 1
Solution
(i) 32n × 153n ÷ 5n = 32n × 33n × 53n ÷ 5n (Recall that (a × b)n = an × bn.)
= 35n × 52n
(ii)
25 × 5 n – 2
52 × 5n – 2
= n –1
(5n –1 is a common factor in the denominator.)
n
n –1
5 –5
5 (5 –1)
5n
=
4(5 n – 1 )
5
=
4
………………………………………………………………………………
Definition of a Surd
2.
A surd is an irrational root of a real number, e.g. 2 and 3 .
Operations on Surds
3.
(a)
a × a = a (b)
(c)
a
=
b
(d)
(e)
a
b
a × b = ab
m a + n a = (m + n) a
m a – n a = (m – n) a
Example 3
Simplify 8 ÷ 2 .
Solution
8÷ 2=
8
2
= 4
=2
Indices, Surds and Logarithms
53
Example 4
Given that a + 2 3 + b 2 = 8 + 5 2, find the possible values of a and of b.
(
)(
)
Solution
(a + 2) (3 + b 2) = 3a + ab
2 + 3 2 + 2b
= 3a + 2b + (ab + 3) 2
By comparing,
3a + 2b = 8 — (1)
ab + 3 = 5 — (2)
(Equate the rational terms and the irrational terms to obtain
2 equations.)
From (2),
ab = 2
2
b = — (3)
a
Substitute (3) into (1):
⎛2 ⎞
3a + 2 ⎜ ⎟ = 8
⎝a ⎠
3a2 – 8a + 4 = 0
(3a – 2)(a – 2) = 0
2
a=
or a = 2
3
b=3
b=1
2
∴ a = , b = 3 or a = 2, b = 1
3
………………………………………………………………………………
Conjugate Surds
4.
a m + b n and a m – b n are conjugate surds.
5.
(a
6.
54
m + b n a m – b n = a2m – b2n, which is a rational number.
)(
)
The product of a pair of conjugate surds is always a rational number.
UNIT 4
Example 5
Simplify
3+3 2
(
)(
3−3 2 .
)
Solution
(
3 +3 2
)(
3–3 2 =
) ( 3)
2
–9
( 2)
2
= 3 – 9(2)
= –15
………………………………………………………………………………
Rationalising the Denominator
7.
To rationalise the denominator of a surd is to make the denominator a rational
number.
b
b
a
(a)
=
×
a
a
a
=
(b)
ab
a
1
=
a+ b
1
×
a+ b
a– b
a–b
=
a– b
a– b
Example 6
Rationalise the denominator of
3 2 +2
.
3 2+3
Solution
3 2 +2
3 2 +2
3 2 −3
=
×
3 2+3
3 2+3
3 2 −3
3 2 3 2 +6 2 −9 2 −6
=
2
3 2 − 32
(
)(
)
(
12 − 3 2
9
4− 2
=
3
)
=
Indices, Surds and Logarithms
55
Example 7
⎛
4 ⎞
The sides of rectangle ABCD are 3 + 8 cm and ⎜5 –
⎟ cm in length.
⎝
2 ⎠
Express, in the form of a + b 2, where a and b are integers,
(i) the area of the rectangle in cm2,
(ii) the area of a square in cm2, given that AC is one of its sides.
(
)
A
B
cm
D
C
cm
Solution
⎛
4 ⎞
(i) Area of rectangle ABCD = 3 + 8 ⎜5 –
⎟
⎝
2 ⎠
(
12
+ 5 8 – 4 4 (Rationalise the
2
12
denominator of
.)
2
12 2
= 15 –
+5 2 2 –8
2
= 15 –
(
(
)
Area of square = AC 2
2
2
⎛
4 ⎞
= 3 + 8 + ⎜5 –
(Apply Pythagoras’ Theorem.)
⎟
⎝
2 ⎠
40
= 9 + 6 8 + 8 + 25 –
+8
2
= 50 + 12 2 – 20 2
(
)
= 50 – 8 2 cm 2
(
56
UNIT 4
)
= 7 + 4 2 cm2
(ii)
)
)
Common Logarithms and Natural Logarithms
8.
log10 is called the common logarithm and it is represented by lg.
9.
loge is called the natural logarithm and it is represented by ln.
Laws of Logarithms
10. (a) loga x + loga y = loga xy
(c) loga xr = r loga x
(b) loga x – loga y = loga
x
y
More Formulae on Logarithms
11. (a)
(b)
(c)
(d)
(e)
log c b
(Change of Base Formula)
log c a
loga 1 = 0
loga a = 1
loga b =
aloga y = y (For logay to be a real number, y > 0)
loga ax = x
Example 8
Simplify log3 81 – log5 125 + log
2
8.
Solution
log3 81 – log5 125 + log
2
8 = log3 34 – log5 53 +
log 2 8
log 2 2
= 4 log3 3 – 3 log5 5 +
=4–3+6
=7
3log 2 2
1
log 2 2
2
Indices, Surds and Logarithms
57
Solving Exponential Equations
12. Given ax = b,
• If b can be expressed as a power of a, e.g. b = ay, then ax = ay ⇒ x = y.
• If b cannot be expressed as a power of a,
lgb
• take common logarithms on both sides, i.e. x lg a = lg b ⇒ x =
, or
lg a
• take natural logarithms on both sides if a = e, i.e. x ln e = ln b ⇒ x = ln b
Example 9
Solve the exponential equation 9 = 34x.
Solution
9 = 34x
32 = 34x
2 = 4x
1
x=
2
Example 10
Solve the equation 3ey – 5 = 2e–y.
Solution
3ey – 5 = 2e–y
2
3ey – 5 – y = 0
e
3(ey)2 – 5ey – 2 = 0 (Multiply by ey.)
Let w = ey.
3w2 – 5w – 2 = 0
(3w + 1)(w – 2) = 0
1
w=–
or
w=2
3
1
ey = –
ey = 2 (Note that ey > 0.)
3
(no solution)
y = ln 2
= 0.693 (to 3 s.f.)
58
UNIT 4
13. Given p(a2x ) + q(ax ) + r = 0,
Step 1: Substitute u = ax to get a quadratic equation pu2 + qu + r = 0.
Step 2: Solve for u and deduce the value(s) of x.
Example 11
Solve the exponential equation 22x + 1 = 6(2x) – 4.
Solution
22x + 1 = 6(2x) – 4
(2 ) (2) = 6(2x) – 4
x 2
Let 2x = y.
2y2 = 6y – 4
2y – 6y + 4 = 0
y2 – 3y + 2 = 0
(y – 2)(y – 1) = 0
y=2
or
y=1
2x = 21
2x = 20
x = 1
x=0
2
Example 12
Without using a calculator, solve the equation 9x ––
28 x
(3 ) + 3 = 0.
3
Solution
28 x
(3 ) + 3 = 0
3
3(9x) – 28(3x) + 9 = 0
3(3x)2 – 28(3x) + 9 = 0
9x –
(Ensure that the exponential terms have the same base.)
Let y = 3x.
3y2 – 28y + 9 = 0
(3y – 1)(y – 9) = 0 (Factorise the quadratic expression.)
1
y=
or
y=9
3
1
3x =
3x = 9
3
x = –1
or
x=2
Indices, Surds and Logarithms
59
Solving Logarithmic Equations
14. To solve logarithmic equations,
Step 1: Change the bases of the logarithmic functions to the same base. We usually choose the smaller as the final base.
Step 2: Use one of the following methods to solve the equations.
(a) If loga x = loga y, then x = y and vice versa.
(b) If loga x = b, then x = ab.
(c) Use the laws of logarithms to combine the terms into the forms
described in method (a) or (b).
………………………………………………………………………………
Example 13
Solve the equation loga 32x – loga (2x2 + x – 54) = 3 loga 2.
Solution
loga 32x – loga (2x2 + x – 54) = 3 loga 2
32x
loga 2
= loga 23
2x + x – 54
32x
=8
2x 2 + x – 54
32x = 16x2 + 8x – 432
16x2 – 24x – 432 = 0
2x2 – 3x – 54 = 0
(2x + 9)(x – 6) = 0
9
x = – (rejected) or x = 6
2
60
UNIT 4
(Substitute your answers
into the original equation
to check if any solution
needs to be rejected.)
Example 14
Solve the equation log3 (x + 2) = 5.
Solution
log3 (x + 2) = 5
x + 2 = 35
x = 241
Example 15
(a) Solve the equation lg (6x + 4) – lg (x – 6) = 1.
(b) Find the value of x given that ex – e = 10.
Solution
(a) lg (6x + 4) – lg (x – 6) = 1
6x + 4
lg
=1
x–6
6x + 4
= 10 (Change to the exponential form.)
x–6
6x + 4 = 10x – 60
4x = 64
x = 16
x–e
(b) e = 10
x – e = ln 10 (Use ln instead of lg because ln e = 1.)
x = e + ln 10
= 5.02 (to 3 s.f.)
Indices, Surds and Logarithms
61
Example 16
Solve the simultaneous equations
e e x = e2 y ,
log4 (x + 2) = 1 + log2 y.
Solution
e e x = e 2 y — (1)
log4 (x + 2) = 1 + log2 y — (2)
From (1),
x
e1e 2 = e 2 y
1+
e
1+
x
2
= e2 y
x
= 2y
2
x = 4y – 2 — (3)
From (2),
log 2 (x + 2)
= 1 + log2 y (Apply the Change of Base Formula.)
log 2 4
log2 (x + 2) = 2 + 2 log2 y (Rearrange the logarithmic terms to one
log2 (x + 2) – log2 y2 = 2
side of the equation.)
x+2
log2
=2
y2
x+2
=4
y2
x = 4y2 – 2
Substitute (3) into (4):
4y – 2 = 4y2 – 2
4y2 – 4y = 0
4y(y – 1) = 0
y=0
or y = 1
x = –2 (rejected) x = 2
∴ x = 2, y = 1
62
UNIT 4
— (4)
(Substitute your answers into the original
equations to check if any solutions need to be
rejected.)
Example 17
At the beginning of 1980, the number of mice in a colony was estimated at 50 000.
The number increased so that, after n years, the number would be 50 000 × e0.05n.
Estimate
(i) the population of the mice, correct to the nearest thousand, at the beginning of
the year 2000;
(ii) the year during which the population would first exceed 100 000.
Solution
(i) At the beginning of year 2000, n = 20.
∴ Population of the mice = 50 000 × e0.05(20)
≈ 136 000 (to the nearest thousand)
(ii) Let 50 000 × e0.05n = 100 000
e0.05n = 2
0.05n = ln 2
n = 13.86
∴ The population will exceed 100 000 in the year 1993.
………………………………………………………………………………
Graphs of Exponential Functions
15. Graphs of y = ax
y
0
y
1
y = ax, where a > 1
x
1
0
y = ax, where 0 < a < 1
x
The graph of y = ax must pass through the point (0, 1) because a0 = 1.
Indices, Surds and Logarithms
63
Graphs of Logarithmic Functions
16. Graphs of y = loga x
y
y
y = loga x,
where a . 1
x
0
1
0
1
Example 18
Sketch the graph of each of the following functions.
(a) y = ex – 1 + 1
(c) y = ln (2x – 3)
(b) y = 2e1 – 3x
(d) y = ln (5 – 3x)
Solution
(a)
y
y = ex – 1 + 1
1
+1
e
O
64
UNIT 4
y=1
x
x
y = loga x,
where 0 , a , 1
y
(b)
2e
y = 2e1 – 3x
x
O
(c)
y
y = ln (2x – 3)
O
x
2
x=
(d)
3
2
y
ln 5
O
y = ln (5 – 3x)
x
4
3
x= 5
3
Indices, Surds and Logarithms
65
Example 19
Sketch the graph of y = ex + 1. By drawing a suitable straight line on the same graph,
find the number of solutions of the equation x + 1 = ln (5 – 2x).
Solution
x + 1 = ln (5 – 2x)
ex + 1 = 5 – 2x
Draw y = 5 – 2x.
y
y = ex + 1
5
e
O
∴ There is 1 solution.
66
UNIT 4
5
2
y = 5 – 2x
x
UNIT
Coordinate Geometry
5
y
B(x2, y2)
M
A(x1, y1)
θ
O
x
Distance between 2 Points
1.
Length of AB = (x2 – x1 )2 + (y2 – y1 )2
Midpoint of 2 Points
2.
⎛ x1 + x2 y1 + y2 ⎞
,
Midpoint of AB = ⎜
⎟
2 ⎠
⎝ 2
Gradient of Line and Collinear Points
3.
Gradient of AB, m =
y2 – y1
y – y2
= tan θ
= 1
x2 – x1
x1 – x2
Coordinate Geometry
67
4.
m>0
i.e. 0° < θ < 90°
(line slopes upwards)
m<0
i.e. 90° < θ < 180°
(line slopes downwards)
y
y
θ
x
O
O
m=0
i.e. θ = 0°
(horizontal line)
x
m is undefined
i.e. θ = 90°
(vertical line)
y
y
x
O
5.
θ
O
x
If A(x1, y1), B(x2, y2) and C(x3, y3) are collinear, then gradient of AB = gradient of
BC = gradient of AC and area of ΔABC = 0.
Parallel and Perpendicular Lines
6.
Given that two lines l1 and l2 have gradients m1 and m2 respectively,
• l1 is parallel to l2 if m1 = m2;
• l1 is perpendicular to l2 if m1m2 = –1.
7.
The perpendicular bisector of a line AB is defined as a line passing through the
midpoint of AB, cutting it into two equal halves and it is also perpendicular to AB.
Perpendicular bisector
A
68
UNIT 5
B
Equation of a Straight Line
8.
Gradient form: y = mx + c, where m is the gradient and c is the y-intercept
9.
Intercept form:
x y
+ = 1, where a and b are the intercepts the line makes on the
a b
x-axis and y-axis respectively
10. General form: Ax + By + C = 0, where A, B and C are constants
………………………………………………………………………………
Example 1
Find the equation of the perpendicular bisector of AB, where A is (3, 10) and B is
(7, 2).
Solution
⎛ 3 + 7 10 + 2 ⎞
,
Midpoint of AB = ⎜
⎟ (The perpendicular bisector of AB passes
2 ⎠
⎝ 2
through the midpoint of AB.)
= (5, 6) 10 – 2
Gradient of AB =
3– 7
= –2
1
Gradient of perpendicular bisector =
(m1m2 = –1)
2
Equation of perpendicular bisector:
1
y – 6 = (x – 5) (To use y – y1 = m(x – x1), we require the gradient and the
2
coordinates of a point on the line.)
1
7
y= x+
2
2
Coordinate Geometry
69
Example 2
A line segment joins P(5, 7) and Q(x, y). The midpoint of the line segment is (4, 2).
Find the coordinates of Q and the equation of the perpendicular bisector of PQ.
Solution
⎛5 + x 7 + y ⎞
,
⎜
⎟ = (4, 2)
2 ⎠
⎝ 2
5+x
7+y
= 4
=2
2
2
5 + x = 8
7+y=4
x = 3
y = –3
∴ Q(3, –3)
7 – (–3)
5–3
=5
1
Gradient of perpendicular bisector = –
5
Equation of perpendicular bisector:
y–2 11
=–
x– 4 55
5y – 10 = –x + 4
14 1
y=
– x
5 5
Gradient of PQ =
Collinear Points
11.
B
C
A
• From the diagram, three points A, B and C lie on the same line. We can say
that they are collinear.
• To show that the points are collinear, determine 2 of the 3 gradients of the line
segments AB, AC and BC. The gradients must be equal, i.e. mAB = mAC,
mAC = mBC or mAB = mBC.
70
UNIT 5
Area of Polygons
12. If A(x1, y1), B(x2, y2), C(x3, y3), … , and N(xn, yn) form a polygon, then
Area of polygon =
1 x1 x2 x3 … xn x1
2 y1 y2 y3 … yn y1
1
(x y + x2y3 + … + xny1 – x2y1 – x3y2 – … – x1yn)
2 1 2
13. If A(x1, y1), B(x2, y2) and C(x3, y3) form a triangle ABC, then
=
Area of ΔABC =
1 x1 x2 x3 x1
2 y1 y2 y3 y1
14. Vertices must be taken in a cyclic and anticlockwise order.
Example 3
Find the area of a triangle with coordinates A(20, –1), B(30, 0) and C(10, 5).
y
(Use a diagram to visualise the relative
positions of the vertices of the triangle.)
C(10, 5)
B(30, 0)
O
A(20, –1)
x
Solution
Area of triangle ABC =
1
2
30 10
20
30
0
–1
0
5
1
(150 – 10) – (100 – 30)
2
1
= (140 – 70)
2
= 35 units2
=
Coordinate Geometry
71
Example 4
A triangle has vertices A(4, 0), B(10, 4) and C(9, 0). Given that ABCD is a
parallelogram, find
(i) the coordinates of the point D,
(ii) the area of the parallelogram ABCD.
Solution
y
4
(Use a sketch to help you
visualise the position of D.)
B
A
C
O
4
9 10
x
D
(i) Let the coordinates of D be (x, y).
Midpoint of BD = Midpoint of AC
⎛10 + x 4 + y ⎞ ⎛ 4 + 9 0 + 0 ⎞
⎜
,
,
⎟ = ⎜
⎟
2 ⎠ ⎝ 2
2 ⎠
⎝ 2
10 + x 4 + 9 4 + y 0 + 0
2 = 2 , 2 = 2
x=3
y = –4
∴ D(3, – 4)
(ii) Area of parallelogram ABCD
=
1 4 3 9 10
2
0 –4 0 4
4
0
1
(–16 + 36 + 36 – 16)
2
= 20 units2
=
72
UNIT 5
(The diagonals of a parallelogram
bisect each other.)
(Remember to take the vertices in a
cyclic and anticlockwise order.)
Example 5
A(–1, –1), B(–2, 2) and C(2, 1) are three vertices of a parallelogram ABCD.
Find the midpoint of AC. Hence, find the coordinates of D.
Solution
Let the coordinates of D be (h, k).
y
B(–2, 2)
C(2, 1)
O
x
A(–1, –1)
(Use a sketch to help you
visualise the position of D.)
D(h, k)
⎛ –1 + 2 –1 + 1⎞
,
Midpoint of AC = ⎜
⎟
2 ⎠
⎝ 2
⎛1 ⎞
= ⎜ , 0⎟
⎝2 ⎠
Midpoint of AC = Midpoint of BD
⎛ –2 + h 2 + k ⎞ ⎛1 ⎞
,
⎜
⎟ = ⎜ , 0⎟
2 ⎠ ⎝2 ⎠
⎝ 2
–2 + h 1 2 + k
= ,
=0
2
2 2
h=3
k = –2
∴ D(3, –2)
Coordinate Geometry
73
Ratio Theorem
15.
Internal point of division
Let the point P divide the line AB
internally in the ratio m : n, then P is
⎛
⎞
the point ⎜nx1 + mx2 , ny1 + my2 ⎟ .
m
+
n
m
+
n
⎝
⎠
m
A (x1, y1)
74
UNIT 5
Let the point Q divide the line AB
externally in the ratio m : n, then Q is
⎛
⎞
the point ⎜ mx2 – nx1 , my2 – ny1 ⎟ .
m
–
n
m
–
n
⎝
⎠
y
y
O
External point of division
n
m
B (x2, y2)
n
B (x2, y2)
P
Q
A (x1, y1)
x
O
x
Example 6
y
C
A(–2, 4)
P(0, 2)
O
Q(4, 1)
x
B
The diagram shows a triangle ABC in which A is the point (–2, 4). The side AB
cuts the y-axis at P(0, 2). The point Q(4, 1) lies on BC and the line AQ is
perpendicular to BC. Find
(i) the equation of BC,
(ii) the coordinates of B.
Given further that Q divides BC internally in the ratio 1 : 3, find
(iii) the coordinates of C,
(iv) the area of triangle ABC.
Solution
4 –1
1
=–
–2 – 4
2
1
=–
2
(i) Gradient of AQ =
Gradient of BC = 2
Equation of BC: y – 1 = 2(x – 4)
y = 2x – 7 – (1)
(To use y – y1 = m(x – x1), we require
the gradient and the coordinates of a
point on the line.)
Coordinate Geometry
75
4–2
= –1
–2 – 0
= –1
(ii) Gradient of AB =
Equation of AB: y = –x + 2 – (2) (We can use y = mx + c because we know
that the y-intercept is 2.)
(Since B lies on AB and BC, we solve the equations of
these 2 lines simultaneously.)
Solving (1) and (2),
x = 3
y = –1
∴ B(3, –1)
(iii)
C(x, y)
3
(Use a sketch to help you visualise the
position of C.)
Q(4, 1)
1
B(3, –1)
Let the coordinates of C be (x, y).
Using Ratio Theorem,
⎛ 3(3) + 1(x) 3(–1) + 1(y)⎞
,
⎟ = (4, 1)
⎜
3+1
3 + 1 ⎠
⎝
9 + x
y–3
= 4,
=1
4
4
x=7
∴ C(7, 7)
y=7
(iv) Area of ∆ABC
=
1
2
–2
3
4 –1
7 –2
7
4
(Remember to take the vertices in a cyclic
and anticlockwise order.)
1
(2 + 21 + 28 – 12 + 7 + 14)
2
= 30 units2
=
76
UNIT 5
UNIT
6
Further Coordinate
Geometry
Equation of a Circle
1. Standard form
(x – a)2 + (y – b)2 = r2
where (a, b) is the centre and r is the radius
2. General form
x2 + y2 + 2gx + 2fy + c = 0
where (–g, –f ) is the centre and g 2 + f 2 – c is the radius
Example 1
Find the equation of the circle with centre (1, 2) and radius of 8.
Solution
Equation of circle:
(x – 1)2 + (y – 2)2 = 82
(x – 1)2 + (y – 2)2 = 64
Further Coordinate Geometry
77
Example 2
A circle has centre (–1, 1) and passes through (2, 5).
(i) Find the equation of the circle.
(ii) Determine if (3, 3) lies on the circumference of the circle.
Solution
(i) Radius, r = (2 + 1)2 + (5 – 1)2 (To find the equation of the circle, we need
the radius and the coordinates of the centre.)
=5
∴ Equation of circle is (x + 1)2 + (y – 1)2 = 25
x2 + 2x + 1 + y2 – 2y + 1 = 25
x2 + y2 + 2x – 2y – 23 = 0
(ii) Substitute x = 3, y = 3 into (x + 1)2 + (y – 1)2:
(3 + 1)2 + (3 – 1)2 = 20
∴ (3, 3) does not lie on the circle. (In fact, (3, 3) lies inside the circle.)
Example 3
A circle has the equation x2 + y2 – 10x + 6y + 9 = 0.
Find the coordinates of the centre and radius of the circle.
Solution
x2 + y2 – 10x + 6y + 9 = 0
(x – 5) – 25 + (y + 3)2 – 9 + 9 = 0
(x – 5)2 + (y + 3)2 = 52
2
Coordinates of centre = (5, –3), radius = 5
78
UNIT 6
Example 4
Find the coordinates of the centre and the radius of a circle with the equation
x2 + y2 – 2x – 4y + 5 = 64.
Solution
x2 + y2 – 2x – 4y + 5 = 64
x2 + y2 – 2x – 4y – 59 = 0
Comparing this with x2 + y2 + 2gx + 2fy + c = 0
2g = –2
2f = –4
c = –59
g = –1
f = –2
Centre of circle (–g, –f) = (1, 2)
Radius of circle = g 2 + f 2 – c
= (–1)2 + (–2)2 + 59
=8
Example 5
Find the radius and the coordinates of the centre of the circle
2x2 + 2y2 – 3x + 4y + 1 = 0.
Solution
2x2 + 2y2 – 3x + 4y + 1 = 0
3
1
x2 + y2 – x + 2y +
=0
2
2
2
⎛
3⎞
9
1
2
⎜x – ⎟ − + (y + 1) – 1 + = 0
4 ⎠ 16
2
⎝
⎛
⎜x –
⎝
∴ Radius =
2
3⎞
17
2
⎟ + (y + 1) =
4 ⎠
16
⎛ 3
⎞
17
, coordinates of centre = ⎜ , – 1⎟
4
⎝ 4
⎠
Further Coordinate Geometry
79
Example 6
Show that the line 4y = x – 3 touches the circle x2 + y2 – 4x – 8y + 3 = 0.
Hence, find the coordinates of the point of contact.
Solution
4y = x – 3 — (1)
x + y – 4x – 8y + 3 = 0 — (2)
2
2
From (1),
x–3
4
Substitute (3) into (2):
y=
— (3)
2
⎛ x – 3⎞
⎛ x – 3⎞
x2 + ⎜
⎟ – 4x – 8 ⎜
⎟ + 3 = 0
⎝ 4 ⎠
⎝ 4 ⎠
x 2 – 6x + 9
– 4x – 2x + 6 + 3 = 0
16
16x2 + x2 – 6x + 9 – 64x – 32x + 96 + 48 = 0
17x2 – 102x + 153 = 0
x2 – 6x + 9 = 0
x2 +
Discriminant = (–6)2 – 4(1)(9)
=0
∴ The line is a tangent to the circle.
Solving x2 – 6x + 9 = 0,
(x – 3)2 = 0
x=3
y=0
∴ Point of contact is (3, 0)
80
UNIT 6
Further Graphs
3.
Graphs of the form y2 = kx, where k is a real number
(a) k . 0
(b) k , 0
y
y
y = kx, k , 0
2
y = kx, k . 0
2
O
4.
x
O
x
Graphs of y = axn
(a) n is even and a . 0
(b) n is even and a , 0
2
e.g. y = 3x e.g. y = –3x2
y
y
x
O
x
O
(c) n is odd and a . 0
(d) n is odd and a , 0
e.g. y = 2x3 e.g. y = –2x3
y
y
O
x
O
x
Further Coordinate Geometry
81
5.
Graphs of y = ax –n
(a) n is even and a . 0
(b) n is even and a , 0
–2
e.g. y = 3x e.g. y = –3x–2
y
y
x
O
x
O
(c) n is odd and a . 0
(d) n is odd and a , 0
e.g. y = 3x–3 e.g. y = –3x–3
y
y
O
82
UNIT 6
x
O
x
6.
1
Graphs of y = ax n
(a) n is even and a . 0
(b) n is even and a , 0
1
6
e.g. y = 4 x y
x
O
(c) n is odd and a . 0
1
e.g. y = –4 x 6
y
x
O
(d) n is odd and a , 0
1
7
1
e.g. y = 3x e.g. y = –3x 7
y
y
O
x
O
x
Further Coordinate Geometry
83
7.
Graphs of y = ax
–
1
n
(a) n is even and a . 0
e.g. y = 4 x
–
1
10
(b) n is even and a , 0
y
e.g. y = – 4 x
–
1
10
y
x
O
(c) n is odd and a . 0
e.g. y = 0.5x
–
1
7
x O
(d) n is odd and a , 0
e.g. y = –0.5x
y
–
1
7
y
84
UNIT 6
O
x
O
x
UNIT
7
Linear Law
(not included for NA)
The Linear Law
1.
y
y = mx + c
gradient = m
(0, c)
O
x
If the variables x and y are related by the equation y = mx + c, then a graph of the
values of y plotted against their respective values of x is a straight line graph.
The straight line has a gradient m and it cuts the vertical axis at the point (0, c).
2.
Linear Law is used to reduce non-linear functions to the linear form y = mx + c.
3.
To reduce confusion, we sometimes denote the horizontal axis as X and the vertical
axis as Y, i.e. Y = mX + c.
Linear Law
85
4.
Some of the common functions and their corresponding X and Y are shown in the
table.
Function
X
Y
1. y = ax + b
x
y
a
++ bb
xn
1
xn
y
xn
1
y
x
y
x
y x
x2
1
x2
x2y
x
n
2. y =
n
3.
1
nn
= xax
++b b
y
4.
y = an x + b
5.
b
b
y = a yx=+a x + i.e y i.e
x =y axx += bax + b
x
x
6.
xy =
a
+ bx
x
7. x = bxy + ay
n
i.e. x2y = bx2 + a
or y =
a b
+ =n
x y
or ay + bx = nxy
9. y = ax2 + bx + n
86
UNIT 7
x
= bx + a
y
y
1 a
= +b
y x
1
x
x
y
1
y
1 ⎛ a ⎞ 1 n
= ⎜– ⎟ +
y ⎝ b ⎠ x b
⎛a ⎞ y b
i.e. y = ⎜ ⎟ +
⎝n ⎠ x n
1
x
1
y
i.e.
or
8.
a
+b
x2
i.e.
i.e.
y–n
= ax + b
x
y
x
x
y
y–n
x
Function
X
Y
10. y = a2x2 + 2abx + b2 i.e. y = ax + b
x
y
or y = –ax – b
x
y
1 1
b
= x–
y a
a
x
1
y
11. y =
a
x–b
i.e.
12. y = axb
i.e. lg y = b lg x + lg a
or ln y = b ln x + ln a
lg x
ln x
lg y
ln y
13. y = axb + n
i.e. lg (y – n) = b lg x + lg a
lg x
lg (y – n)
14. y = abx
i.e. lg y = x lg b + lg a
x
lg y
⎛lgb ⎞ lg a
i.e. lg y = x ⎜ ⎟ –
n
⎝ n ⎠
x
lg y
16. yax = b + n
i.e. lg y = (–lg a)x + lg (b + n)
x
lg y
17. yb = 102x + a
i.e. lg y =
2
a
x+
b
b
x
lg y
18. yb = e2x + a
i.e. ln y =
2
a
x+
b
b
x
ln y
15. y n =
bx
a
Linear Law
87
Example 1
The variables x and y are related in such a way that when y – 3x is plotted against
x2, a straight line passing through (2, 1) and (5, 7) is obtained. Find
(i) y in terms of x,
(ii) the values of x when y = 62.
Solution
With reference to the sketch graph and using Y to represent y – 3x and X to represent x2,
Y –1 7 –1
the equation of the straight line is
=
.
X–2 5–2
i.e. Y – 1 = 2(X – 2)
Y = 2X – 3
y – 3x
(i) y – 3x = 2x2 – 3
y = 2x2 + 3x – 3
(5, 7)
(ii) When y = 62,
2x2 + 3x – 3 = 62
2x2 + 3x – 65 = 0
(x – 5)(2x + 13) = 0
13
x = 5 or x = –
2
(2, 1)
O
88
UNIT 7
x2
Example 2
It is known that x and y are related by the formula xy = a + bx, where a and b are
constants.
x
2
4
6
8
10
y
38
21.3
15.8
13.1
11.5
Express this equation in a form suitable for drawing a straight line graph. Draw this
graph for the given data and use it to estimate the value of a and of b.
Solution
Since xy = a + bx,
a
y=
+ b.
x
1
Plot y against , gradient = a, vertical-axis intercept = b.
x
x
2
4
6
8
10
y
38
21.3
15.8
13.1
11.5
1
x
0.50
0.25
0.17
0.13
0.10
Linear Law
89
y
40
(0.5, 38)
30
20
10
0
(0.08, 10)
0.1
0.2
0.3
From the graph, vertical-axis intercept = 4.8
38 – 10
Using (0.08, 10) and (0.5, 38), gradient =
0.5 – 0.08
= 66.7 (to 3 s.f.)
∴ a = 66.7, b = 4.8
90
UNIT 7
0.4
0.5
1
x
Example 3
The volume (V) of a container and the height of the container (x) are connected by
an equation of the form V = hkx, where h and k are constants.
x
V
1
12.6
2
25.1
3
50.1
4
90.0
5
199.5
(a) Express V = hkx in a form suitable for drawing a straight line graph.
(b) Plot this straight line graph and use it to estimate the value of h and of k.
Solution
(a)
V = hkx
lg V = (lg k)x + lg h
Y = mX + c
Gradient = lg k
Y-intercept = lg h
Linear Law
91
(b)
x
1
V
2
12.6
lg V
3
25.1
1.10
4
50.1
1.40
5
90.0
1.70
199.5
1.95
2.30
lg V
3
(4.4, 2.1)
2
1
(0, 0.8)
0
1
2
3
From the graph,
vertical-axis intercept = 0.8
lg h = 0.8
h = 6.31 (to 3 s.f.)
Using (0, 0.8) and (4.4, 2.1),
2.1 – 0.8
gradient =
4.4 – 0
= 0.295 (to 3 s.f.)
lg k = 0.295
k = 1.97 (to 3 s.f.)
92
UNIT 7
4
5
x
UNIT
8
Trigonometric Functions
and Equations
Basic Trigonometric Ratios
1.
90° – θ
h
O
•
•
•
θ
y
h
x
cos θ =
h
y
tan θ =
x
sin θ =
y
x
1
h
•
cosec θ = sin θ = y
•
sec θ = cos θ = x
•
cot θ = tan θ = y
•
•
•
cos (90°− θ) = sin θ
cot (90°− θ) = tan θ
cosec (90° − θ) = sec θ
1
h
1
x
Complementary Angles
2.
•
•
•
sin (90°− θ) = cos θ
tan (90°− θ) = cot θ
sec (90°− θ) = cosec θ
Trigonometric Functions and Equations
93
Basic Angle (or Reference Angle)
3.
The basic angle, α, is the acute angle between a rotating radius about the origin and
the x-axis.
y
α
α
•
•
•
α
α
x
sin (–θ) = –sin θ
cos (–θ) = cos θ
tan (–θ) = –tan θ
Signs of Trigonometric Ratios in the Four Quadrants
4.
Sine positive
All positive
x = 180° – α
x = 180° + α
x=α
α
α
Tangent positive
94
UNIT 8
α
α
x = 360° – α
Cosine positive
Example 1
4
Given that cos θ = – and that 180° , θ , 270°, find the value of sin θ and
5
tan θ.
Solution
y
a
–4
θ
x
5
θ lies in the 3rd quadrant.
(–4)2 + a2 = 52
a2 = 25 – 16
=9
a = –3 (Since a lies in the negative y-axis, a , 0.)
3
sin θ = –
5
3
tan θ =
4
Trigonometric Functions and Equations
95
Example 2
Given that sec α =
17
and that α is an acute angle, find the value of each of the
15
following.
(i) sin α
(ii) tan (90° – α)
(iii) cos (180° – α)
Solution
sec α =
1
17
15
i.e. cos α =
(Recall that sec x =
.)
cos x
15
17
y
17
α
15
(i) sin α =
8
17
(ii) tan (90° – α) =
15
8
(iii) cos (180° – α) = – cos α
15
=–
17
96
UNIT 8
(Use Pythagoras’ Theorem to find the length
of the side opposite α.)
90° – α 8
x
Example 3
4
Given that 270° < β , 360° and sin β = – , find the value of each of the
5
following without using a calculator.
(i) cos β
(ii) tan β
Solution
β lies in the 4th quadrant.
42 + a2 = 52
a2 = 25 – 16
=9
a = 3 (Since a lies in the
positive x-axis, a . 0.) 3
(i) cos β =
5
4
(ii) tan β = –
3 y
β
a
5
x
–4
Trigonometric Functions and Equations
97
Trigonometric Ratios of Special Angles
5.
θ
sin θ
cos θ
tan θ
0°
0
1
0
1
2
3
2
3
3
30° =
45° =
π
4
2
2
2
2
1
60° =
π
3
3
2
1
2
3
90° =
π
2
1
0
Undefined
180° = π
0
–1
0
3π
2
–1
0
Undefined
360° = 2π
0
1
0
270° =
98
UNIT 8
π
6
Example 4
Find all the angles between 0° and 360° inclusive which satisfy each of the following
equations.
(i) 5 sin2 x – 6 sin x cos x = 0
⎛ 3y
⎞
(ii) 1 + 2 sin ⎜ + 15°⎟ = 0
⎝ 2
⎠
Solution
(i) 5 sin2 x – 6 sin x cos x = 0 (Do not make the mistake of dividing throughout
sin x (5 sin x – 6 cos x) = 0 by sin x, as you will then be short of answers.)
5 sin x – 6 cos x = 0 sin x = 0
5 sin x = 6 cos x
x = 0°, 180°, 360°
sin x
6
(Recall that tan x =
.)
tan x =
5
cos x
α = 50.19° (to 2 d.p.)
x = 50.2°, 230.2° (to 1 d.p.) ∴ x = 0°, 50.2°, 180°, 230.2°, 360°
⎛ 3y
⎞
(ii) 1 + 2 sin ⎜ + 15°⎟ = 0
⎝ 2
⎠
⎛ 3y
⎞
1
sin ⎜ + 15°⎟ = –
2
⎝ 2
⎠
α = 30°
3y
+ 15° = 210°, 330° (The required angles are in the 3rd and
2
4th quadrants.)
3y
= 195°, 315° 2
∴ y = 130°, 210°
Trigonometric Functions and Equations
99
Graphs of Trigonometric Functions
6.
y = a sin bx
• amplitude = a
360°
• period =
b
y
a
y = a sin bx
x
0
7.
y = a cos bx
• amplitude = a
360°
• period =
b
–a
y
y = a cos bx
a
x
0
–a
8.
y = a tan bx
180°
• period =
b
9.
y
y = a tan bx
x
0
To sketch the graphs of y = a sin bx + c or y = a cos bx + c or y = a tan bx + c:
Step 1: Draw the graph of y = a sin bx or y = a cos bx or y = a tan bx.
Step 2: If c . 0, shift the graph up by c units.
If c , 0, shift the graph down by c units.
||
100
UNIT 8
Example 5
Sketch the graph of y = 3 sin 2x – 1 in the domain 0° < x < 180°.
Solution
First, sketch y = 3 sin 2x.
It has an amplitude of 3 and period of 180°.
y
3
y = 3 sin 2x – 1
0
–1
45°
90°
135°
180°
135°
180°
x
–3
Next we shift the graph down by 1 unit.
y
y = 3 sin 2x – 1
2
0
–1
45°
90°
x
–4
Trigonometric Functions and Equations
101
Example 6
Sketch the graph of y = tan x + 1 for 0° < x < 360°.
Solution
First, sketch y = tan x .
y
y = tan x
0
90°
180°
270°
360°
x
Next, shift the graph up by 1 unit.
y
y = tan x + 1
1
0
102
UNIT 8
90°
180°
270°
360°
x
Example 7
Sketch on the same diagram, for 0  x  2π, the graphs of
(i) y = 1 + 4 sin x,
(ii) y = 2 cos x.
Hence, deduce the number of roots of the equation 2 cos x = 1 + 4 sin x for
0  x  2π.
Solution
y
(To draw y = 1 + 4 sin x, we first draw y = 4 sin x, before
translating it by 1 unit upwards.)
5
y = 2 cos x
2
1
0
π
2
–2
–3
π
3π
2
2π
x
y = 1 + 4 sin x
From the graph, there are 2 roots. (The question is asking for the number of
intersection points between the graphs.)
Fundamental Identities
10.
sin
tan
cos
1
sec
sin θ
cos θ
cos θ
12. cot θ =
sin θ
1
13. sec θ =
cos θ
1
14. cosec θ =
sin θ
cot
cosec
11. tan θ =
Trigonometric Functions and Equations
103
UNIT
9
Trigonometric Identities
and Formulae
Fundamental Identities
1.
sin2 A + cos2 A = 1
3.
cot2 A + 1 = cosec2 A
2.
tan2 A + 1 = sec2 A
Example 1
Prove that
1 + sin x
= tan x + cot x + sec x.
sin x cos x
Solution
RHS = tan x + cot x + sec x
sin x cos x
1
=
+
+
cos x sin x cos x
sin 2 x + cos 2 x + sin x
(Use the identity sin2 x + cos2 x = 1.)
sin x cos x
1 + sin x
=
= LHS (proven)
sin x cos x
=
104
UNIT 9
Example 2
Show that
1
1
+
= 2 cosec2 θ.
1 + cos θ 1 – cos θ
Solution
1
1
+
1 + cos θ 1 – cos θ
1 – cos θ + 1 + cos θ
=
1 – cos 2 θ
2
=
(Use the identity sin2 θ + cos2 θ = 1.)
sin 2 θ
LHS =
= 2 cosec2 θ = RHS (proven)
(Recall that cosec θ =
1
.)
sin θ
Example 3
Prove that sec4 θ – sec2 θ = tan2 θ + tan4 θ.
Solution
LHS = sec4 θ – sec2 θ
= sec2 θ (sec2 θ – 1)
= (1 + tan2 θ)(tan2 θ) (Use the identity tan2 θ + 1 = sec2 θ.)
= tan2 θ + tan4 θ = RHS (proven)
Trigonometric Identities and Formulae
105
Example 4
Find all the angles between 0° and 360° inclusive which satisfy the equation
3 tan2 y + 5 = 7 sec y.
Solution
3 tan2 y + 5 = 7 sec y
3(sec2 y – 1) + 5 = 7 sec y (Use tan2 y + 1 = sec2 y to obtain a
3 sec2 y – 7 sec y + 2 = 0
quadratic equation in sec y.)
(3 sec y – 1)(sec y – 2) = 0
3 sec y – 1 = 0
or
sec y – 2 = 0
1
sec y =
sec y = 2
3
1
cos y = 3 (no solution)
cos y =
2
α = 60°
∴ y = 60°, 300°
………………………………………………………………………………
Compound Angle Formulae
4.
sin (A ± B) = sin A cos B ± cos A sin B
5.
cos (A ± B) = cos A cos B sin A sin B
tan A ± tan B
tan (A ± B) =
1  tan A tan B
6.
106
UNIT 9
Example 5
Find all the angles between 0° and 360° which satisfy the equation
5 sin (x + 60°) = cos (x – 30°).
Solution
5 sin (x + 60°) = cos (x – 30°)
5[sin x cos 60° + cos x sin 60°] = cos x cos 30° + sin x sin 30°
⎡1
⎤
3
3
1
5 ⎢ sin x +
cos x⎥ =
cos x + sin x
2
2
2
2
⎣
⎦
5 sin x + 5 3 cos x = 3 cos x + sin x
4 sin x = –4 3 cos x
sin x
tan x = – 3 (Recall that tan x =
.)
cos x
α = 60°
∴ x = 120°, 300° (x lies in the 2nd and 4th quadrants.)
………………………………………………………………………………
Special Identities
7.
sin (θ + 2nπ) = sin θ, where n is an integer
9.
tan (θ + 2nπ) = tan θ, where n is an integer
8.
cos (θ + 2nπ) = cos θ, where n is an integer
10. sin (90° ± θ) = cos θ
11. cos (90° ± θ) = sin θ
12. tan (90° ± θ) = cot θ
13. sin (180° ± θ) = sin θ
14. cos (180° ± θ) = –cos θ
15. tan (180° ± θ) = ± tan θ
Trigonometric Identities and Formulae
107
Double Angle Formulae
16. sin 2A = 2 sin A cos A
17. cos 2A = cos2 A − sin2 A
= 2 cos2 A − 1
= 1 − 2 sin2 A
2 tan A
18. tan 2A =
1 – tan 2 A
Example 6
4
Given that tan θ = – and 270° < θ < 360°, find the value of
3
(i) cos (–θ ),
(ii) cos (90° – θ ),
(iii) sin (180° + θ ),
(iv) sin 2θ.
Solution
(i) cos (–θ) = cos θ
3
=
5
(ii) cos (90° – θ) = sin θ
4
=–
5
(iii) sin (180° + θ ) = –sin θ
4
=
5
(iv) sin 2θ = 2 sin θ cos θ
⎛ 4 ⎞⎛ 3⎞
= 2 ⎜– ⎟⎜ ⎟
⎝ 5 ⎠⎝5 ⎠
24
=–
25
108
UNIT 9
y
θ
3
O
5
x
–4
Example 7
Given that cos 2x =
(i) cos x,
(ii) sin x.
127
and 270° < 2x < 360°, find the value of
162
Solution
(i) Since 270° < 2x < 360°, then 135° < x < 180°.
∴ x lies in the 2nd quadrant.
127
cos 2x =
162
127
2
2 cos x –1 =
162
289
2
2 cos x =
162
289
cos2 x =
324
289
cos x = ±
324
17
=±
18
17
∴ cos x = –
(cos x , 0 since x lies in the 2nd quadrant.)
18
(ii)
35
18
sin x =
y
18
x
–17
O
x
35
18
Trigonometric Identities and Formulae
109
Half Angle Formulae
A
in the Double Angle Formulae.
2
A
A
19. sin A = 2 sin
cos
2
2
2 A
2 A
20. cos A = cos
sin
2
2
A
= 2 cos2
–1
2
A
= 1 – 2 sin2
2
A
2 tan
2
21. tan A =
2 A
1 – tan
2
Replace A with
R-Formulae
22. a sin θ + b cos θ = R sin (θ + α)
23. a sin θ − b cos θ = R sin (θ − α)
24. a cos θ + b sin θ = R cos (θ − α)
where R = a 2 + b 2 and tan α =
25. a cos θ − b sin θ = R cos (θ + α)
26. For the expression a sin θ ± b cos θ or a cos θ ± b sin θ,
• Maximum value = R
• Minimum value = –R
110
UNIT 9
b
a
Example 8
Using the R-formula, find the maximum and minimum values of 6 sin x – 5 cos x
for values of x, where 0° , x , 360°.
Solution
6 sin x – 5 cos x = R sin (x – α)
R = 6 2 + 5 2 = 61
⎛5 ⎞
α = tan–1 ⎜ ⎟
⎝6 ⎠
= 39.8°
∴ 6 sin x – 5 cos x = 61 sin (x – 39.8°)
Minimum value = – 61 (when sin (x – 39.8°) = –1)
Maximum value = 61 (when sin (x – 39.8°) = 1)
Example 9
Solve 3 sin 2x + 2 sin x = 0 for 0° < x < 360°.
Solution
3 sin 2x + 2 sin x = 0
3(2 sin x cos x) + 2 sin x = 0
3 sin x cos x + sin x = 0
sin x (3 cos x + 1) = 0
sin x = 0
α = 70.53°
x = 109.5°, 250.5°
or
x = 0°, 180°, 360°
3 cos x + 1 = 0
cos x = –
1
3
(The required angles are in the 2nd and 4th quadrants.)
Trigonometric Identities and Formulae
111
Example 10
π
,
2
(i) obtain the maximum value of 4 cos x − 3 sin x + 5 and the corresponding value
of x,
(ii) solve the equation 4 cos x − 3 sin x = 2.5 for values of x between 0 and 2π
inclusive.
By expressing 4 cos x − 3 sin x in the form R cos (x + α), where R . 0 and 0 , α <
Solution
4 cos x – 3 sin x = R cos (x + α)
R = 4 2 + 32 = 5
3
tan α =
4
α = 0.6435 (to 4 s.f.)
∴ 4 cos x – 3 sin x = 5 cos (x + 0.644)
(i) Maximum value = 5 + 5 (Maximum value of 4 cos x – 3 sin x is 5)
= 10
Maximum value occurs when cos (x + 0.6435) = 1,
i.e. x + 0.6435 = 2π
x = 5.64 (to 3 s.f.)
(ii) 4 cos x – 3 sin x = 2.5
5 cos (x + 0.6435) = 2.5 (Use the expression obtained earlier to solve the
cos (x + 0.6435) = 0.5 equation.)
π
α=
3
x + 0.6435 = 1.047, 5.235 (to 4 s.f.) (x + 0.6435 lies in the 1st and
x = 0.404, 4.59 (to 3 s.f.)
4th quadrants.)
112
UNIT 9
UNIT
Proofs in
Plane Geometry
10
(not included for NA)
Useful Properties and Concepts that are learnt in
O Level Mathematics
1.
Angle Properties
(a) Alternate angles between parallel lines are equal
P
c
a
R
b
Q
d
S
Since PQ // RS, a = d and b = c
(b)
Corresponding angles between parallel lines are equal
c
P
a
R
d
b
Q
S
Since PQ // RS, a = c and b = d
(c)
Interior angles between parallel lines are supplementary
P
R
a
c
b
Q
d
S
Since PQ // RS, a + c = 180° and b + d = 180°
Proofs in Plane Geometry
113
2.
Properties of Congruent Triangles
• Corresponding sides are equal in length.
• Corresponding angles are equal.
3.
Congruence Tests for Triangles
(i) SSS
AB = XY, AC = XZ and BC = YZ
X
B
A
Y
(ii) SAS
AB = XY, AC = XZ and A = X
C
Z
X
B
A
Y
C
Z
(iii) AAS or ASA
AB = XY, A = X and B = Y
X
A
B
Y
C
(iv) RHS
Only applicable for right-angled triangles.
BC = YZ, AB = XY and C = Z = 90°
Z
Z
A
Y
C
114
UNIT 10
B
X
4.
Properties of Similar Triangles
• All corresponding angles are equal.
• All corresponding sides are proportional in length.
AE AB EB
=
=
AD AC DC
2
Area of ΔAEB ⎛ AE ⎞
= ⎜ ⎟
Area of ΔADC ⎝ AD ⎠
5.
A
E
B
D
C
Similarity Tests for Triangles
(i) AA
A = X and B = Y
X
B
A
Y
C
Z
(ii) SSS
AB BC AC
=
=
XY YZ XZ
X
B
Y
A
C
Z
(iii) SAS
AB BC
and B = Y
=
XY YZ
X
B
Y
A
C
Z
Proofs in Plane Geometry
115
6.
Circles
(a)  at centre = 2 at circumference
An angle at the centre is twice any angle at the circumference subtended by
the same arc, i.e. a = 2b. O
b
a
(b) Rt.  in a semicircle
Every angle at the circumference subtended by the diameter of a circle is a
right angle, i.e. a = 90º. a
O
(c) s in the same segment
Angles in the same segment of a circle are equal, i.e. a = b.
Or
If AB = BC, then ADB = BDC.
a
D
b
A
C
B
(d) s in opp. segments are supplementary
In a cyclic quadrilateral, the opposite angles are supplementary,
i.e. a + c = 180º and b + d = 180°.
a
b
116
UNIT 10
c
d
(e) ⊥ bisector of a chord passes through the centre of the circle
A straight line drawn from the centre to bisect a chord is perpendicular to the
chord, i.e. OC ⊥ AB ⇔ AC = BC.
O
A
B
C
(f) Equal chords are equidistant from the centre
Chords which are equidistant from the centre are equal, i.e. AB = DE ⇔ OC = OF.
(ΔOAB  ΔODE)
E
F
D
O
A
B
C
(g) Tangent ⊥ radius
A tangent to a circle is perpendicular to the radius
drawn to the point of contact,
i.e. OC ⊥ PQ.
O
P
Q
C
(h) Tangents from an external point
(i) Tangents drawn to a circle from an external point are equal, i.e. PA = PB.
(ii) The line joining the external point to the centre of the circle bisects the angle
between the tangents,
i.e. APO = BPO and AOP = BOP.
A
(ΔOAP  ΔOBP)
P
O
B
Proofs in Plane Geometry
117
7. Midpoint Theorem for Triangles
In ∆ABC, if D and E are the midpoints of the sides AB and AC respectively, then
DE // BC and DE =
1
BC.
2
A
E
D
C
B
8.
Tangent-chord Theorem (Alternate Segment Theorem)
The angle between a tangent and a chord meeting the tangent at the point of contact
is equal to the inscribed angle on the opposite side of the chord,
i.e. ∠BAE = ∠BCA and ∠CAD = ∠CBA.
C
B
D
118
UNIT 10
A
E
Example 1
The diagram shows a circle, centre O, with diameter AC and AB = AD. AC and BD
intersect at X.
(a) Prove that ∆ABC and ∆ADC are congruent.
(b) Prove that BD is perpendicular to AC.
B
A
O
X
C
D
Solution
(a) AB = AD
AC is a common side for the two triangles. (The hypotenuse of both
∠ABC = ∠ADC = 90° (rt. ∠ in a semicircle) triangles are the same.)
ΔABC is congruent to ΔADC (RHS congruence).
(b) Since ΔABC is congruent to ΔADC and they share the same base (AC),
BX = DX.
Since AC passes through the centre of the circle, AC ⊥ BD (⊥ bisector of a
chord passes through the centre of the circle).
Proofs in Plane Geometry
119
Example 2
In the figure, BD and FE are tangents to the circle, centre O. BED is a tangent to
the circle at B and ACD is a straight line. ∠CED = 90°.
A
F
C
O
B
E
Prove that
(i) ∠ABC = ∠ECD,
(ii) ∆ABD is similar to ∆BCD.
Solution
(i) ∠ECD = ∠ACF (vert. opp. ∠s)
∠ACF = ∠ABC (∠s in alt. segments)
∠ABC = ∠ECD
(ii) In ∆ABD and ∆BCD,
∠BAD = ∠CBD (∠s in alt. segments)
∠ABD = 90° (Tangent ⊥ radius)
∠BCD = ∠BCA = 90° (rt. ∠ in a semicircle)
i.e. ∠ABD = ∠BCD
∆ABD is similar to ∆BCD (AA Similarity Test).
120
UNIT 10
D
UNIT
11
Differentiation and its
Applications
Formulae
1.
2.
3.
d n
(x ) = nxn – 1
dx
d
(axn) = anxn – 1
dx
d
(k) = 0
dx
Addition/Subtraction Rules
4.
If y = u(x) ± v(x),
dy d
d
= [u(x)] ± [v(x)]
dx dx
dx
Example 1
Differentiate 2x3 – 8x2 +
1
– 4 with respect to x.
x2
Solution
⎞
d ⎛ 3
1
2
⎜2x – 8x + 2 – 4⎟
dx ⎝
⎠
x
d
=
(2x3 – 8x2 + x–2 – 4)
dx
= 6x2 – 16x – 2x–3
2
= 6x2 – 16x – 3
x
(Change
1
to x–2.)
x2
Differentiation and its Applications
121
Chain Rule
dy dy du
=
×
dx du dx
5.
If y is a function of u, then
6.
d ⎡
(ax + b)n⎤⎦ = an(ax + b)n – 1
dx ⎣
7.
In general,
d
[f(x)]n = n[f(x)]n – 1 × f ´(x)
dx
Example 2
Differentiate 5 – 4 x 2 with respect to x.
Solution
d
dx
(
5 – 4 x2
)
1
= d 5 – 4 x2 2
dx
1
–
1
= 5 – 4 x 2 2 (–8x)
2
4x
= –
5 – 4 x2
(
(
)
)
(Chain Rule)
………………………………………………………………………………
Product Rule
8.
122
If y = uv, where u and v are functions of x, then
UNIT 11
dy
dv
du
=u ×v
dx
dx
dx
Example 3
Differentiate 2x(3x3 – 2)3 with respect to x.
Solution
d ⎡
2x(3x 3 – 2)3⎤⎦
dx ⎣
= 2x(3)(3x3 – 2)2(9x2) + 2(3x3 – 2)3 (Product Rule and Chain Rule)
= 2(3x3 – 2)2(27x3 + 3x3 – 2) (Take out common factors.)
= 2(3x3 – 2)2(30x3 – 2)
= 4(3x3 – 2)2(15x3 – 1)
…………………………………………………………………………
Quotient Rule
9.
du
dv
dy v dx – u dx
u
=
If y = , where u and v are functions of x, then
dx
v
v2
Example 4
Differentiate
3x 2 + 4
with respect to x.
2x + 5
Solution
2
d ⎡ 3x + 4 ⎤
⎢
⎥
dx ⎢⎣ 2x + 5 ⎥⎦
–
⎛ 1⎞
6x 2x + 5 – (3x 2 + 4) ⎜ ⎟ (2x + 5) 2 (2)
2
⎝ ⎠
=
2x + 5
1
=
6x(2x + 5) – (3x 2 + 4)
=
9x 2 + 30x – 4
(Quotient Rule)
3
(2x + 5) 2
(2x + 5)3
Differentiation and its Applications
123
Equations of Tangent and Normal to a Curve
10. Equation of a straight line: y – y1 = m(x – x1)
y
tangent
(x1, y1)
normal
11. To find the equation of a tangent, we need:
dy
Gradient of tangent, m =
dx
Coordinates of a point that lies on the tangent, (x1, y1)
12. To find the equation of a normal, we need:
dy
Gradient of tangent =
dx
dy
Gradient of normal = –1 ÷
dx
Coordinates of a point that lies on the normal, (x1, y1)
124
UNIT 11
x
Example 5
A curve has the equation y = x2 + 3x.
(i) Find the equation of the tangent to the curve at (1, 4).
(ii) Find the equation of the normal to the curve at (1, 4).
Solution
dy
(i) Step 1: Find
.
dx
dy
= 2x + 3
dx
Step 2: Substitute x = 1 into
dy
= 2(1) + 3
dx
=5
dy
to find the gradient of the tangent.
dx
Step 3: Find the equation of the tangent.
y – 4 = 5(x – 1)
y – 4 = 5x – 5
y = 5x – 1
(ii) Step 1: Find the gradient of the normal.
1
Gradient of normal = –
Gradient of tangent
1
= –
5
Step 2: Find the equation of the normal.
1
y – 4 = – (x – 1)
5
5y – 20 = – x + 1
5y = –x + 21
Differentiation and its Applications
125
Example 6
The equation of a curve is y =
5
. Find
1 – 3x
dy
,
dx
(ii) the equation of the tangent to the curve at x = 2,
(iii) the equation of the normal to the curve at x = 2.
(i)
Solution
5
1 – 3x
dy (1 – 3x)(0) – 5(–3)
=
dx
(1 – 3x)2
15
=
(1 – 3x)2
(i)
y=
(ii) When x = 2,
y = –1
dy 3
=
dx 5
3
(x – 2)
5
3
11
y= x–
5
5
∴ Equation of tangent: y + 1 =
(iii) Gradient of normal = –
5
3
(m1m2 = –1)
5
(x – 2)
3
5
7
y= – x+
3
3
∴ Equation of normal: y + 1 = –
126
UNIT 11
Connected Rates of Change
dx
is the rate of change of x with respect to time t and y = f(x), then the rate of
dt
dy dy dx
=
×
change of y with respect to t is given by
.
dt dx dt
13. If
14. A positive rate of change is an increase in the magnitude of the quantity involved
as the time increases.
15. A negative rate of change is a decrease in the magnitude of the quantity involved
as the time increases.
Example 7
x
. Find the rate of
3x + 7
change of x at the instant when x = 1, given that y is changing at a rate of 3.5 units/s at
this instant.
Two variables, x and y, are related by the equation y =
Solution
x
3x + 7
dy (3x + 7)(1) – x(3)
=
dx
(3x + 7)2
3x + 7 – 3x
=
(3x + 7)2
7
=
(3x + 7)2
y=
dy dy dx
=
×
,
dt dx dt
7
dx
×
3.5 =
(3 + 7)2 dt
dx
= 50 units/s
dt
Using
Differentiation and its Applications
127
Example 8
A cube has sides of x cm. Its volume, V cm3, is expanding at a rate of 30 cm3/s.
Find the rate of change of x of the cube when the volume is 64 cm3.
Solution
V = x3
dV
= 3x2
dx
When V = 64, x = 4.
When x = 4,
dV
= 3(4)2
dx
= 48
dV
Given that
= 30,
dt
dV dV dx
=
×
dt
dx dt
dx
30 = 48 ×
dt
dx
= 0.625 cm/s
dt
128
UNIT 11
UNIT
12
Further Applications
of Differentiation
Increasing/Decreasing Functions
1.
2.
Function in x
y
f(x)
First derivative
dy
dx
f´(x)
Second derivative
d2 y
dx 2
f ´´(x)
Third derivative
d3y
dx 3
f´´´(x)
If y is an increasing function (y increases as x increases), the gradient is positive,
i.e.
e.g.
dy
. 0.
dx
y
y
y = f(x)
y = g(x)
x
y
x
y
y = k(x)
y = h(x)
x
x
Further Applications of Differentiation
129
Example 1
Find the set of values of x for which f(x) = 2x3 – 10x2 + 14x + 5 is an increasing
function.
Solution
f(x) = 2x3 – 10x2 + 14x + 5
f ´(x) = 6x2 – 20x + 14
When f ´(x) . 0,
6x2 – 20x + 14 . 0
3x2 – 10x + 7 . 0
(3x – 7)(x – 1) . 0
3.
x,1
or
x.
7
3
x
7
3
1
If y is a decreasing function (y decreases as x increases), the gradient is negative,
dy
i.e.
, 0.
dx
y
y
y = f(x)
y = g(x)
x
y
y
y = k(x)
y = h(x)
x
130
UNIT 12
x
x
Example 2
Find the set of values of x for which y =
1 3 5 2
x – x + 6x is a decreasing function.
3
2
Solution
y=
1 3 5 2
x – x + 6x
3
2
dy
= x2 – 5x + 6
dx
For y to be a decreasing function,
x2 – 5x + 6 , 0
(x – 3)(x – 2) , 0
2
dy
, 0.
dx
3
x
∴2,x,3
Further Applications of Differentiation
131
Stationary Points
4.
dy
If a point (x0, y0) is a stationary point of the curve y = f(x), then
= 0 when
dx
x = x0, i.e. the gradient of the tangent at x = x0 is zero.
5.
A stationary point can be a maximum point, a minimum point or a point of inflexion.
Determining the Nature of Stationary Points
6.
First Derivative Test: Use
Maximum point
x
x0
x
dy
dx
.0
0
,0
slope
/
–
\
–
stationary
point
Point of inflexion
x0
x
dy
dx
.0
0
.0
slope
/
–
/
stationary
point
UNIT 12
Minimum point
x–
x0
x+
dy
dx
,0
0
.0
slope
\
–
/
+
x
–
132
dy
.
dx
stationary
point
Point of inflexion
x–
x0
x+
dy
dx
,0
0
,0
slope
\
–
\
+
stationary
point
7.
d2 y
.
dx 2
d2 y
If
, 0, the stationary point is a maximum point.
dx 2
d2 y
If
. 0, the stationary point is a minimum point.
dx 2
Second Derivative Test: Use
•
•
d2 y
= 0, the stationary point can be a maximum point, a minimum
dx 2
point or a point of inflexion. Use the First Derivative Test to determine
the nature.
•
If
Problems on Maxima and Minima
8. Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
Step 6:
Find a relationship between the quantity to be maximised or minimised
and the variable(s) involved.
If there is more than one variable involved, use substitution to reduce it to
one independent variable only.
Find the first derivative of the expression obtained above.
Equate the first derivative to zero to obtain the value(s) of the variable.
Check the nature of the stationary point.
Find the required maximum or minimum value of the quantity.
Further Applications of Differentiation
133
Example 3
A curve has the equation y = 3(x + 1)2. Find the coordinates of the stationary point
and deduce the nature of the stationary point.
Solution
dy
= 6(x + 1)
dx
dy
Let
= 0,
dx
6(x + 1) = 0
x = –1
When x = –1, y = 0.
To find the nature of the stationary point, we perform the First Derivative Test.
x
–1.1
–1
–0.9
dy
dx
,0
0
.0
slope
stationary point
(–1, 0) is a minimum point.
134
UNIT 12
Example 4
16
and that z = x2 + 2y. Given that x is positive, find the
x4
value of x and of y that makes z a stationary value and show that in this case, z
has a minimum value.
It is given that y =
Solution
16
– (1)
x4
z = x2 + 2y – (2)
y=
Substitute (1) into (2):
32
z = x2 + 4
x
= x2 + 32x–4
dz
= 2x – 128x–5
dx
128
= 2x – 5
x
dz
When
= 0,
dx
128
2x – 5 = 0
x
128
2x = 5
x
x6 = 64
x=±2
(Express z in terms of one variable.)
Given that x is positive,
x=2
y = 1 (Substitute x = 2 into (1) to obtain the value of y.)
d2 z
= 2 + 640x–6 (Use the Second Derivative Test to show that z has a
dx 2
minimum value.)
640
=2+ 6
x
When x = 2,
d2 z
= 12 . 0
dx 2
∴ z has a minimum value.
Further Applications of Differentiation
135
Example 5
r cm
x cm
The diagram shows a rectangle of length x cm and 2 semicircles each of radius r cm.
The perimeter of the figure is 400 cm and the area of the rectangle is A cm2.
(a) Show that A = 400r – 2πr2.
dA
(b) Find an expression for
.
dr
(c) Calculate
(i) the value of r for which A is a maximum,
(ii) the maximum value of A.
Solution
(a) Given that the perimeter is 400 cm,
2x + 2πr = 400
(As A is expressed in terms of r only, we make use of
x = 200 – πr the perimeter to obtain an equation involving x and r,
A = x(2r)
before substituting it into A.)
= 2r(200 – πr)
= 400r – 2πr2 (proven)
dA
(b)
= 400 – 4πr
dr
136
UNIT 12
dA
= 0,
dr
400 – 4πr = 0
100
r=
π
2
d A
2 = –4π , 0 (Use the Second Derivative Test to check
dr
that A is a maximum.)
∴ A is a maximum.
(c) (i) When
100
,
π
2
⎛100 ⎞
⎛100 ⎞
A = 400 ⎜ ⎟ – 2π ⎜ ⎟
⎝ π ⎠
⎝ π ⎠
40 000 20 000
=
–
π
π
20 000
=
π
= 6370 (to 3 s.f.)
∴ The maximum value of A is 6370 (to 3 s.f.).
(ii) When r =
Further Applications of Differentiation
137
Example 6
A cylinder, which is made using a thin sheet of metal, has a volume of 500 cm3, radius
of x cm and height of h cm.
(a) Express h in terms of x and hence, express the total surface area, A cm2, in terms
of x.
(b) Find the value of x for which A will be a minimum.
x
h
Solution
(a)
V = πx2h
2
πx h = 500
500
h=
π x2
⎛500 ⎞
A = 2πx2 + 2πx ⎜ 2 ⎟
⎝ π x ⎠
= 2πx2 + 1000
x
(b)
A = 2πx2 + 1000x–1
dA
= 4πx – 1000x–2
dx
1000
= 4πx –
x2
To find the minimum value of A,
4πx –
138
UNIT 12
1000
=0
x2
250
x3 =
π
250
x= 3
π
= 4.30 (to 3 s.f.)
dA
= 0.
dx
UNIT
13
Differentiation of Trigonometric,
Logarithmic & Exponential
Functions and their Applications
(not included for NA)
Differentiation of Trigonometric Functions
1.
2.
3.
Ensure that your calculator is in the radian mode.
d
d
(sin x) = cos x (cos x) = –sin x
dx
dx
d
d
(tan x) = sec2 x
(sec x) = sec x tan x
dx
dx
d
d
(cot x) = –cosec2 x (cosec x) = cosec x cot x
dx
dx
d
[sin (Ax + B)] = A cos (Ax + B)
dx
d
[cos (Ax + B)] = –A sin (Ax + B)
dx
d
[tan (Ax + B)] = A sec2 (Ax + B)
dx
Example 1
Differentiate each of the following with respect to x.
(a) 3 sin (2x + 1)
(b) (2x + 1) cos 3x
(c) x3 tan (3x + 2)
Solution
d
(a)
[3 sin (2x + 1)] = 3[2 cos (2x + 1)]
dx
= 6 cos (2x + 1)
d
(b)
(2x + 1) cos 3x = (2x + 1)(3)(–sin 3x) + cos 3x (2) (Product Rule)
dx
= –3(2x + 1) sin 3x + 2 cos 3x
d 3
⎡x tan (3x + 2)⎤⎦ = x3(3) sec2 (3x + 2) + tan (3x + 2) (3x2) (Product Rule)
(c)
dx ⎣
= 3x2 [x sec2 (3x + 2) + tan (3x + 2)]
Differentiation of Trigonometric, Logarithmic & Exponential Functions and their Applications
139
Example 2
Find the gradient of the curve y = x sin x at the point where x = 1.
Solution
y = x sin x
dy
= x cos x + sin x
dx
(Gradient of curve refers to
dy
.)
dx
When x = 1,
dy
= cos 1 + sin 1 (Radian mode)
dx
= 1.38 (to 3 s.f.)
∴ Gradient of curve at x = 1 is 1.38
………………………………………………………………………………
4.
5.
d ⎡ n ⎤
sin x⎦ = n sinn – 1 x cos x
dx ⎣
d ⎡ n ⎤
cos x⎦ = –n cosn – 1 x sin x
dx ⎣
d ⎡ n ⎤
tan x⎦ = n tann – 1 x sec2 x
dx ⎣
d ⎡ n
n–1
sin (Ax + B)⎤⎦ = An sin (Ax + B) cos (Ax + B)
dx ⎣
d ⎡ n
n–1
cos (Ax + B)⎤⎦ = –An cos (Ax + B) sin (Ax + B)
dx ⎣
d
⎡tan n (Ax + B)⎤⎦ = An tann – 1 (Ax + B) sec2 (Ax + B)
dx ⎣
In general,
d ⎡ n
d
6.
sin f(x)⎤⎦ = n sin n – 1 f(x) × [sin f(x)]
dx ⎣
dx
d ⎡ n
d
cos f(x)⎤⎦ = n cos n – 1 f(x) × [cos f(x)]
dx ⎣
dx
d ⎡ n
d
tan f(x)⎤⎦ = n tan n – 1 f(x) × [tan f(x)]
dx ⎣
dx
140
UNIT 13
Example 3
Differentiate each of the following with respect to x.
(a) cos2 (1 – 3x)
(b) 3 tan3 (2x – π)
(c) sin2 (3x + 2) cos x2
Solution
d ⎡ 2
cos (1 – 3x)⎤⎦ = 2 cos (1 – 3x) [–(–3) sin (1 – 3x)]
dx ⎣
= 6 cos (1 – 3x) sin (1 – 3x)
(a)
(b)
d ⎡
3 tan 3 (2x – π)⎤⎦ = 3⎡⎣(3) tan 2 (2x – π)⎤⎦⎡⎣(2) sec 2 (2x – π)⎤⎦ (Chain Rule)
dx ⎣
= 18 tan2 (2x – π) sec2 (2x – π)
d
[sin2 (3x + 2) cos x2]
dx
= (2) sin (3x + 2) (3) cos (3x + 2) (cos x2) + sin2 (3x + 2) (2x) (–sin x2)
= 6 sin (3x + 2) cos (3x + 2) cos x2 – 2x sin2 (3x + 2) sin x2 (Product Rule and
Chain Rule)
(c)
………………………………………………………………………………
Differentiation of Logarithmic Functions
d
1
(ln x) =
dx
x
d
a
8.
[ln (ax + b)] = ax + b
dx
d
9. In general, d [ ln f (x)] = f ′ (x) , where f ´(x) =
[f(x)] .
dx
dx
f(x)
10. As far as possible, make use of the laws of logarithms to simplify logarithmic
expressions before finding the derivatives.
7.
Differentiation of Trigonometric, Logarithmic & Exponential Functions and their Applications
141
Example 4
Differentiate each of the following with respect to x.
(a) ln (3x + 1)
(b) ln (2x2 + 5)3
⎛ 2x ⎞
⎛
⎞
(c) ln ⎜ 2
(d) ln ⎜8 + 4 x ⎟
⎟ +
4
3x
3x
–
5
⎠
⎝
⎝
⎠
(e) ln [x (5x3 – 2)]
(f) x3 ln (4x – 1)
Solution
d
3
[ln (3x + 1)] = 3x + 1
dx
d ⎡
d
(b)
ln (2x 2 + 5)3⎤⎦ = ⎡⎣3 ln (2x 2 + 5)⎤⎦
dx ⎣
dx
⎛ 4 x ⎞
= 3⎜ 2
⎟ (Power Law of Logarithms)
⎝2x + 5 ⎠
12x
= 2
2x + 5
(a)
d ⎡ ⎛ 2x ⎞⎤ d ⎡
⎢ln ⎜
ln 2x – ln (3x 2 + 4)⎤⎦
(c)
⎟⎥ =
dx ⎢⎣ ⎝ 3x 2 + 4 ⎠⎥⎦ dx ⎣
6x
2
–
=
2x 3x 2 + 4
1
6x
= – 2
x 3x + 4
d ⎡ ⎛ 8 + 4 x ⎞⎤ d
(d) dx ⎢ln ⎜ 3x – 5 ⎟⎥ = dx [ln (8 + 4 x) – ln (3x – 5)]
⎠⎦
⎣ ⎝
4
3
=
–
8 + 4x 3x – 5
1
3
=
–
2 + x 3x – 5
142
UNIT 13
d ⎡
d
ln x(5x 3 – 2)⎤⎦ = ⎡⎣ln x + ln (5x 3 – 2)⎤⎦
dx ⎣
dx
15x 2
1
= + 3
x 5x – 2
(e)
(f)
dd ⎡⎡ 33
4 ⎞⎞ + 3x 22 ln (4 x – 1)
3 ⎛⎛ 4
xx ln
ln (4
(4xx –– 1)
1)⎤⎦⎤⎦ == xx 3⎜⎜
⎟⎟ + 3x ln (4 x – 1)
dx
4
x
dx⎣⎣
4
x
⎝⎝ –– 11⎠⎠
4 x 33
2
== 4 x ++ 3x
ln (4
(4xx –– 1)
1)
3x 2 ln
44xx –– 11
Example 5
ln x
. Find the rate of
3x + 7
change of x at the instant when x = 1, given that y is changing at a rate of 0.18 units/s
at this instant.
Two variables, x and y, are related by the equation y =
Solution
ln x
3x + 7
y=
⎛1 ⎞
(3x + 7) ⎜ ⎟ – 3 ln x
dy
⎝ x ⎠
=
dx
(3x + 7)2
=
3x + 7 – 3x ln x
x(3x + 7)2
dy dy dx
=
×
,
dt dx dt
3(1) + 7 – 3(1) ln 1 dx
0.18 =
×
dt
1(3 + 7)2
dx
= 1.8 units/s
dt
Using
Differentiation of Trigonometric, Logarithmic & Exponential Functions and their Applications
143
Example 6
ln 2x
. Find the rate of change of y at the
3x 2
instant when y = 0, given that x is changing at a rate of 2 units/s at this instant.
x and y are related by the equation y =
Solution
ln 2x
3x 2
1
= x–2 ln 2x
3
⎛ 2 ⎞
dy 1
1
= (–2)x –3 ln 2x + x –2 ⎜ ⎟ (Product Rule)
dx 3
3 ⎝2x ⎠
2 ln 2x
1
=–
+ 3
3x 3
3x
1 – 2 ln 2x
=
3x 3
y=
When y = 0,
ln 2x = 0
2x = e0
1
x=
2
dy dy dx
Using
=
× ,
dt dx dt
⎡⎡
⎛1 ⎞⎤⎤
22⎜⎛⎜1⎟⎞⎟⎥⎥
⎢⎢11 –– 22 ln
ln
dy
⎝⎝22⎠⎠⎥
dy = ⎢⎢
⎥ × 2
= ⎢
33
dt
⎥⎥ × 2
⎛
⎞
dt ⎢
11 ⎞
⎛
⎢⎢ 33⎜⎜2 ⎟⎟
⎥⎥
⎝⎝2⎠⎠
⎣⎣
⎦⎦
11
=
= 55 3 units/s
3
………………………………………………………………………………
Differentiation of Exponential Functions
d x
(e ) = ex
dx
d ax + b
12.
(e ) = aeax + b
dx
d f(x)
d
13. In general,
(e ) = f ´(x)ef(x), where f ´(x) =
[f(x)].
dx
dx
11.
144
UNIT 13
Example 7
Differentiate each of the following with respect to x.
(a) e2 – 3x
(c)
e3x
x2 + 1
(b) x2e4x
(d)
esin x + 1
e cos x
Solution
d 2 – 3x
(e ) = –3e2 – 3x
dx
d 2 4x
(b)
(x e ) = x2(4e4x) + 2xe4x (Product Rule)
dx
= 2xe4x (2x + 1)
(a)
(c)
2
3x
3x
d ⎡ e 3 x ⎤ (x + 1)(3)e – e (2x)
⎢ 2 ⎥ =
dx ⎣x + 1⎦
(x 2 + 1)2
=
=
(Quotient Rule)
3(x 2 + 1)e 3 x – 2xe 3 x
(x 2 + 1)2
e 3 x ⎡⎣3(x 2 + 1) – 2x⎤⎦
(x 2 + 1)2
cos x
sin x
sin x
⎛ sin x ⎞
+ 1)e cos x (– sin x)
(d) d ⎜e cos+x 1⎟ = e (cos xe ) – (e
dx ⎝ e
(e cos x )2
⎠
e cos x esin x cos x + (esin x + 1)e cos x sin x
=
e 2 cos x
cos x ⎡ sin x
e ⎣e cos x + (esin x + 1)sin x⎤⎦
=
e 2 cos x
sin x
e cos x + (esin x + 1)sin x
=
e cos x
(Quotient Rule)
Differentiation of Trigonometric, Logarithmic & Exponential Functions and their Applications
145
Example 8
The equation of a curve is y = ex cos x, where 0 , x , π.
Find the x-coordinate of the stationary point of the curve.
Solution
y = ex cos x
dy
= ex (–sin x) + cos x (ex)
dx
= ex (cos x – sin x)
dy
When
= 0,
dx
x
e (cos x – sin x) = 0
ex = 0 (no solution)
or
cos x – sin x = 0
cos x = sin x
tan x = 1
π
x=
4
Example 9
1
x
Given that the equation of a curve is y = e 2 +
4
,
1
x
e2
(i) find the coordinates of the stationary point on the curve,
(ii) determine the nature of the stationary point.
Solution
1
(i)
x
y = e2 +
1
x
4
1
e2
x
= e 2 + 4e
1
– x
2
⎛ 1 ⎞ – 1 x
dy 1 12 x
= e + 4 ⎜– ⎟ e 2
dx 2
⎝ 2 ⎠
146
UNIT 13
=
1
– x
1 12 x
e – 2e 2
2
When
dy
= 0,
dx
1
– x
1 12 x
e – 2e 2 = 0
2
1
– x
1 12 x
e = 2e 2
2
1
e2
e
x
1
– x
2
=4
ex = 4
x = ln 4
1
y = e2
ln 4
=2+
=4
+
4
2
4
e
1
ln 4
2
∴ Coordinates of stationary point are (ln 4, 4)
(ii)
⎛ 1 ⎞ – 12 x
d 2 y ⎛1 ⎞⎛1 ⎞ 12 x
=
–
(2)
e
⎜
⎟
⎜
⎟
⎜– ⎟ e
⎝ 2 ⎠
dx 2 ⎝2 ⎠⎝2 ⎠
=
1
– x
1 12 x
e +e 2
4
When x = ln 4,
d2 y
2 = 1 > 0
dx
∴ The stationary point is a minimum.
Differentiation of Trigonometric, Logarithmic & Exponential Functions and their Applications
147
UNIT
Integration
14
Integration
∫
∫
1.
If y = f(x), then y dx = f(x) dx.
2.
If
∫
dy
= g(x), then g(x) dx = y + c, where c is an arbitrary constant.
dx
Formulae and Rules
3.
4.
5.
6.
∫ k dx = kx + c, where k is a constant
∫ ax dx = axn + 1 + c, where n ≠ –1
+ c, where n ≠ –1
∫ (ax + b) dx = (axa(n+ b)
+ 1)
∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx
n +1
n
n +1
n
Example 1
Find
(a)
(c)
(e)
148
∫ 5 dx, ∫ (2x – 3x + 6) dx,
∫ 3(2x – 5) dx.
UNIT 14
3
6
(b)
(d)
∫ 3x dx,
∫ x ⎛⎜⎝3x + 7x ⎞⎟⎠ dx,
5
2
Solution
(a)
(b)
∫ 5 dx = 5x + c
∫ 3x dx = 3x5 + 1
5 +1
5
=
(c)
∫ (2x
3
+c
1 6
x +c
2
2x 3 + 1 3x1 + 1
–
+ 6x + c
3+1
1+1
1
3
= x4 – x2 + 6x + c
2
2
– 3x + 6) dx =
∫ x ⎛⎜⎝3x
∫
7 ⎞
(Multiply x into the terms in the bracket
+ ⎟ dx = (3x3 + 7) dx x ⎠
before doing the integration.)
3+1
3x
=
+ 7x + c
3+1
3 4
= x + 7x + c
4
(d)
2
∫
3(2x – 5)6 + 1
(e) 3(2x – 5)6 dx =
+ c (It is not necessary to find the expansion
(6 + 1)(2)
of (2x – 5)6.)
3
7
=
(2x – 5) + c
14
Example 2
Find the equation of the curve which passes through the point (2, 10) and dy
4
for which
= 3x2 – 2 .
dx
x
Solution
dy
4
= 3x2 – 2
dx
x
= 3x2 – 4x–2
∫
y = (3x2 – 4x–2) dx
= x3 + 4x–1 + c
4
= x3 +
+c
x
Integration
149
When x = 2, y = 10,
4
10 = 23 +
+c
2
c=0
∴ Equation of the curve is y = x3 +
4
x
………………………………………………………………………………
Integration of Trigonometric Functions
7.
8.
9.
10.
11.
12.
150
∫ sin x dx = –cos x + c
∫ cos x dx = sin x + c
∫ sec x dx = tan x + c
∫ sin (Ax + B) dx = – A1 cos (Ax + B) + c
∫ cos (Ax + B) dx = A1 sin (Ax + B) + c
∫ sec (Ax + B) dx = A1 tan (Ax + B) + c
2
2
UNIT 14
Example 3
Find
(a) cos (5x + 3) dx,
(c)
∫
∫ 2 sec
2
(b)
∫ 3 sin (3x – 1) dx,
(8 – 3x) dx.
Solution
(a)
(b)
∫ cos (5x + 3) dx = 15 sin (5x + 3) + c
⎡– cos (3x − 1)⎤
∫ 3 sin (3x – 1) dx = 3⎢⎣ 3 ⎥⎦ + c
∫
= –cos (3x – 1) + c
⎡tan (8 − 3x)⎤
(c) 2 sec2 (8 – 3x) dx = 2 ⎢
⎥ + c
–3
⎣
⎦
2
= – tan (8 – 3x) + c
3
∫
(Note that 2 sec2 (8 – 3x) dx
2
≠ – sec3 (8 – 3x) + c)
9
………………………………………………………………………………
13. Methods of Integrating Trigonometric Functions:
• Use trigonometric identities e.g. 1 + tan2 x = sec2 x
• Use double angle formulae e.g. cos 2x = 2 cos2 x – 1 or cos 2x = 1 – 2 sin2 x
Integration
151
Example 4
Find
(a)
(c)
∫ 4 tan 3x dx,
∫ 6 cos 2x dx.
2
∫ sin x cos x dx,
2
Solution
(a)
(b)
∫ 4 tan
2
∫
3x dx = 4 (sec2 3x – 1) dx
⎡1
⎤
= 4 ⎢ tan 3x – x⎥ + c
3
⎣
⎦
4
= tan 3x – 4 x + c
3
1
(b) sin x cos x dx =
2 sin x cos x dx
2
1
=
sin 2x dx
2
1 ⎡– cos 2x⎤
= ⎢
+c
2 ⎣ 2 ⎥⎦
1
= – cos 2x + c
4
x
x
(c) 6 cos2 dx = 3 2 cos2 dx
2
2
A
= 3 (cos x + 1) dx (cos A = 2 cos2 – 1 )
2
= 3[sin x + x] + c
= 3 sin x + 3x + c
∫
∫
152
UNIT 14
∫
∫
∫
∫
Example 5
Prove that (2 cos θ – sin θ)2 =
∫
3
5
cos 2θ – 2 sin 2θ + .
2
2
Hence, find (2 cos x – sin x)2 dx.
Solution
LHS = (2 cos θ – sin θ)2
= 4 cos2 θ + sin2 θ – 4 sin θ cos θ
⎛1 + cos 2θ ⎞ ⎛1 – cos 2θ ⎞
= 4 ⎜
⎟ + ⎜
⎟ – 2(2 sin θ cos θ )
2
2
⎝
⎠ ⎝
⎠
1 1
= 2 + 2 cos 2θ + – cos 2θ – 2 sin 2θ
2 2
3
= cos 2θ – 2 sin
sin22θθ + 5 = RHS (shown)
2
2
= RHS (shown)
∫ (2 cos x – sin x)
2
dx =
∫ Q 23 cos 2x – 2 sin 2x + 52 R dx
3
sin 2x (–2 cos 2x) 5
2
=
–
+ x+c
2
2
2
3
5
= sin 2x + cos 2x + x + c
4
2
………………………………………………………………………………
Integration of
14.
15.
16.
1
ax + b
∫∫ 1x dx = ln x + c
∫∫ ax1+ b dx = 1a ln ax + b + c
´(x)
ff'(x)
In general, ∫∫
dx = ln f(x) + c
f(x)
f(x)
Integration
153
Example 6
Find
(a)
∫ 3x5+ 5 dx,
(c)
∫ 4 x 2x+ 3x
2
4
3
(b)
∫ 2 –43x dx,
dx.
Solution
(a)
∫ 3x5+ 5 dx = 5∫ 3x1+ 5 dx
=
(b)
∫ 2 –43x dx = 4∫ 2 –13x dx
=
∫ 4 x 2x+ 3x
2
3
154
UNIT 14
∫
4
–3
dx
–3 2 – 3x
4
= – ln (2 – 3x) + c
3
(c)
∫
3
5
dx
3 3x + 5
5
= ln (3x + 5) + c
3
4
∫
⎛ 4 x 2 3x 4 ⎞
dx = ⎜ 3 + 3 ⎟ dx
2x ⎠
⎝ 2x
⎛2 3 ⎞
= ⎜ + x⎟ dx
⎝ x 2 ⎠
∫
= 2 ln x +
3 2
x +c
4
(Manipulate the expression to obtain
f ´(x)
one in the form
.)
f(x)
Example 7
Express
2x + 4
in partial fractions. Hence, find
(x + 1)(x – 2)
∫ (x +2x1)(x+ 4– 2) dx .
Solution
Let
2x + 4
A
B
=
+
.
(x + 1)(x – 2) x + 1 x – 2
By Cover-Up Rule,
2
8
A = – and B =
3
3
2x + 4
2
8
=–
+
(x + 1)(x – 2)
3(x + 1) 3(x – 2)
⎡
2
8 ⎤
2x + 4
+
dx
dx = ⎢–
3(x
+
1)
3(x
– 2)⎥⎦
(x + 1)(x – 2)
⎣
2
8
= – ln (x + 1) + ln (x – 2) + c
3
3
∫
∫
Integration
155
Example 8
Find
+ 15
dx.
∫ (x –x2)(x
+ 3)
Solution
Let
x + 15
A
B
=
+
.
(x – 2)(x + 3) x – 2 x + 3
By Cover-up Rule,
17
12
and B = –
5
5
x + 15
17
12
=
–
(x – 2)(x + 3)
5(x – 2) 5(x + 3)
⎡ 17
12 ⎤
x + 15
–
dx = ⎢
⎥ dx
(x – 2)(x + 3)
⎣ 5(x – 2) 5(x + 3)⎦
A=
∫
∫
=
17
12
ln(x – 2) – ln(x + 3) + c
5
5
………………………………………………………………………………
Integration of ex
17.
18.
156
∫ e dx = e + c
∫ e dx = 1a e
x
ax + b
UNIT 14
x
ax + b
+c
Example 9
Find
(a)
(c)
∫ e dx,
∫ 6e dx, 3x
x
3
(b)
(d)
∫e
∫e
2x + 3
dx,
3x – 1
2e
–4
x
dx.
Solution
∫
∫
1
(a) ∫ e 3 x dx = e 3 x + c
3
1
(b) ∫ e 2 x + 3 dx = e 2 x + 3 + c
2
∫
x
6e 3
+c
(c) ∫ 6e dx =
1
3
x
3
x
(d)
∫
= 18e 3 + c
e 3x – 1 – 4
dx =
2e x
∫
∫
⎛e 3 x – 1
4 ⎞
⎜ x – x ⎟ dx
2e ⎠
⎝ 2e
⎛ 1 2 x – 1
⎞
– 2e – x ⎟ dx
= ⎜ e
⎝ 2
⎠
1 2x – 1
e
2e – x
–
+c
= 2
2
–1
1
2
= e 2 x – 1 + x + c
4
e
Integration
157
UNIT
Applications of Integration
15
Definite Integrals
1.
2.
∫
3.
∫
4.
∫
5.
6.
158
∫
∫
∫
b
a
a
a
b
a
b
a
c
b
f(x) dx = [F(x)]a = F(b) – F(a)
f(x) dx = 0
f(x) dx = –
∫
cf(x) dx = c
∫
f(x) dx =
a
b
a
∫
b
a
a
b
a
f(x) dx
f(x) dx +
f(x) ± g(x) dx =
UNIT 15
f(x) dx
b
∫
b
a
∫
c
b
f(x) dx
f(x) dx ±
b
∫ g(x) dx
a
Example 1
Evaluate
(a)
∫
3
1
⎛ 2 10
⎞
(b)
⎜x – 2 + 3⎟ dx ⎝
⎠
x
Solution
(a)
∫
3
1
⎛ 2 10
⎞
⎜x – 2 + 3⎟ dx =
⎝
⎠
x
∫
3
1
(x
2
∫
4
1
5x – 4 dx
– 10x –2 + 3 dx
)
3
⎡1
⎤
= ⎢ x 3 + 10x –1 + 3x⎥
⎣3
⎦1
3
⎡1
⎤
10
= ⎢ x 3 +
+ 3x⎥
3
x
⎣
⎦1
⎡ 10 ⎤ ⎡1
⎤
= ⎢9 +
+ 9⎥ – ⎢ + 10 + 3⎥
3
⎣
⎦ ⎣3
⎦
=8
(b)
∫
4
1
5x – 4 dx =
∫
4
1
1
(5x – 4) 2 dx
4
⎡
⎤
3
⎢(5x – 4) 2 ⎥
⎥
= ⎢
⎢ ⎛⎜ 3⎞⎟(5) ⎥
⎢⎣ ⎝2 ⎠
⎥⎦
1
=
=
=
4
3 ⎤
2 ⎡
(5x – 4) 2 ⎥
⎢
15 ⎣
⎦1
3 ⎤
2 ⎡ 23
16 – 12 ⎥
⎢
15 ⎣
⎦
42
5
Applications of Integration
159
Example 2
Given that
(i)
(iii)
∫
∫
1
5
4
1
∫
5
1
f(x) dx = 10, find the value of each of the following.
∫
(ii)
f(x) dx ⎡f(x) + 3 x⎤ dx +
⎣
⎦
∫
5
4
5
1
2f(x) dx
f(x) dx
Solution
(i)
∫
(ii)
∫
1
5
5
1
f(x) dx = – 10
2f(x) dx = 2
(iii)
∫
4
1
∫
5
1
= 2(10)
= 20
⎡f(x) + 3 x⎤ dx +
⎣
⎦
∫
5
4
f(x) dx =
∫
4
1
∫
f(x) dx + 3
5
=
= 10 + 2 x 2
= 10 + 2 4 – 12
UNIT 15
1
3
3
2
= 24
∫
4
1
1
x 2 dx +
4
⎡ 3 ⎤
⎢x 2 ⎥
f(x)dx + 3 ⎢ ⎥
3
⎢⎣ ⎥⎦
2 1
160
f(x) dx
4
1
3
∫
5
4
f(x) dx
Example 3
Find
d ⎛ 1 ⎞
⎟ and hence find the value of
⎜
dx ⎝9 – 2x 2 ⎠
∫
2
1
12x
dx .
(9 – 2x 2 )2
Solution
2
d ⎛ 1 ⎞ (9 – 2x )(0) – 1(–4 x)
=
⎟
⎜
dx ⎝9 – 2x 2 ⎠
(9 – 2x 2 )2
=
∫
2
1
4x
(9 – 2x 2 )2
12x
dx = 3
(9 – 2x 2 )2
2
4x
dx (Make use of the answer in the first part
2 2
1 (9 – 2x )
of the question.)
2
⎡ 1 ⎤
= 3⎢
2 ⎥
⎣9 – 2x ⎦1
∫
⎡1 1⎤
= 3⎢ – ⎥
⎣1 7⎦
18
=
7
Example 4
Evaluate each of the following.
(a)
∫
π
3
0
(b)
3 sin 3x dx
∫
π
4
0
(sec 2 x + 2 cos x) dx
Solution
(a)
∫
π
3
0
π
3
⎡– cos 3x⎤
3 sin 3x dx = 3⎢
⎥
⎣ 3 ⎦0
π
= – [cos 3x]03
= – [cos π – cos 0]
= – [–1 – 1]
=2
Applications of Integration
161
∫
(b)
π
4
0
π
(sec 2 x + 2 cos x) dx = [tan x + 2sin x]04
⎡ π
π⎤
= ⎢tan + 2sin ⎥ – [tan 0 + 2sin 0]
4
4⎦
⎣
=1 + 2
………………………………………………………………………………
Area bounded by the x-axis
7.
For a region above the x-axis:
Area bounded by the curve y = f(x), the lines x = a and x = b and the x-axis is
∫
b
a
f(x) dx.
y
O
8.
a
b
x
For a region below the x-axis:
Area bounded by the curve y = f(x), the lines x = a and x = b and the x-axis is
∫
b
a
f(x) dx .
y
O
a
x
b
y = f(x)
162
UNIT 15
Example 5
Find the area of the shaded region bounded by the curve y = 2x3, the x-axis and the
lines x = 1 and x = 3.
y
y = 2x3
1
O
Solution
∫
=∫
Area of shaded region =
3
1
3
1
3
x
y dx
2x 3 dx
3
⎡1 ⎤
= ⎢ x 4 ⎥
⎣2 ⎦1
1 4
[3 – 14 ]
2
2
= 40
1 units
= [34 – 14 ]
2
= 40 units2
=
Applications of Integration
163
Example 6
The figure shows part of the curve y = (x – 2)(x – 5). Find the area of the shaded
region.
y
10
y = (x – 2)(x – 5)
2
O
5
x
Solution
∫
5
2
(x – 2)(x – 5) dx =
∫
5
2
(x 2 – 7x + 10) dx
5
⎡x 3 7x 2
⎤
= ⎢ –
+ 10x⎥
2
⎣ 3
⎦2
⎡5 3 7(5)2
⎤ ⎡2 3 7(2)2
⎤
= ⎢ –
+ 10(5)⎥ – ⎢ –
+ 10(2)⎥
2
2
⎣ 3
⎦ ⎣ 3
⎦
= 4.5 units2
………………………………………………………………………………
9.
For an area enclosed above and below the x-axis:
Area bounded by the curve y = f(x) and the x-axis as shown below is
∫
b
a
f(x) dx +
∫
c
b
f(x) dx
y
y = f(x)
O
164
UNIT 15
a
b
c
x
Example 7
The diagram shows part of the curve y = (x – 1)(x – 2).
Find the area bounded by the curve and the x-axis.
y
2
y = (x – 1)(x – 2)
1
O
Solution
∫
=∫
Area of shaded region =
1
0
1
0
x
2
∫
– 3x + 2) dx + ∫
(x – 1)(x – 2) dx +
(x 2
2
1
2
1
(x – 1)(x – 2) dx
(x 2 – 3x + 2) dx
1
2
⎡1
⎤
⎡1
⎤
3
3
= ⎢ x 3 – x 2 + 2x⎥ + ⎢ x 3 – x 2 + 2x⎥
2
2
⎣3
⎦0 ⎣3
⎦1
⎡5
⎤ 2 5
= ⎢ – 0⎥ + –
⎣6
⎦ 3 6
= 1 unit 2
………………………………………………………………………………
Area bounded by the y -axis
10. For a region on the right side of the y-axis:
Area bounded by the curve x = f(y), the lines y = a and y = b and the y-axis is
∫
b
a
f(y) dy.
y
b
x = f(y)
a
O
x
Applications of Integration
165
Example 8
The figure shows part of the curve y = x3. Find the area of the shaded region.
y
y = x3
27
8
x
O
Solution
y = x3
x=3y
1
= y3
∫
=∫
Area of shaded region =
27
8
27
8
UNIT 15
27
4
27
⎤
( y) ⎥⎦
3
8
3
[81 – 16]
4
= 48.75 units2
=
166
1
y 3 dy
⎡ 4 ⎤
⎢y 3 ⎥
= ⎢ ⎥
4
⎢⎣ ⎥⎦
3 8
⎡ 3
= ⎢
⎣4
x dy
11. For a region on the left side of the y-axis:
Area bounded by the curve x = f(y), the lines y = a and y = b and the y-axis is
=
∫
b
a
f (y) dy .
y
x = f(y)
b
a
O
x
Applications of Integration
167
Example 9
Calculate the area of the shaded region shown in the figure.
y
x = (y – 4)2
(1, 5)
4
(1, 3)
x
O
Solution
∫
Area of (P + Q) =
5
3
(y – 4)2 dy
5
⎡(y – 4) ⎤
= ⎢
⎥
⎣ 3 ⎦3
3
y
=
=
1 ⎛ 1⎞
– ⎜– ⎟
3 ⎝ 3⎠
P
(1, 5)
Q
(1, 3)
4
2
units2
3
x = (y – 4)2
O
Area of shaded region = Area of rectangle – Area of (P + Q)
2
= (1)(2) –
3
4
2
= units
3
168
UNIT 15
x
Example 10
The diagram shows the curve y = x2 – 4. It cuts the line y = 5 at P(3, 5). The line x = 4
intersects the curve at R(4, 16). Find the area of the shaded region PQR.
y
y = x2 – 4
R
P
Q
y=5
x
O
x=4
Solution
∫
= ∫
Area of PQR =
4
3
4
3
⎡(x 2 – 4) – 5⎤ dx
⎣
⎦
(x 2 – 9) dx
4
⎡1
⎤
= ⎢ x 3 – 9x⎥
⎣3
⎦3
⎡ 44
⎤
= ⎢–
– (–18)⎥
3
⎣
⎦
1
2
= 3 units
3
Applications of Integration
169
UNIT
16
Kinematics
(not included for NA)
Relationship between Displacement, Velocity and
Acceleration
1.
Differentiation:
dv d 2 s
=
dt dt 2
ds
dt
displacement, s
Integration:
s=
velocity, v
∫ v dt
acceleration, a
v=
∫ a dt
Common Terms used in Kinematics
170
2.
Displacement, s, is defined as the distance moved by a particle in a specific direction.
3.
Velocity, v, is defined as the rate of change of displacement with respect to time. v can take on positive or negative values.
4.
Acceleration, a, is defined as the rate of change of velocity with respect to time.
a can take on positive or negative values.
When a . 0, acceleration occurs.
When a , 0, deceleration occurs.
5.
Initial At rest Stationary Particle is at the fixed point
Maximum/minimum displacement
Maximum/minimum velocity
UNIT 16
t
v
v
s
v
a
=
=
=
=
=
=
0
0
0
0
0
0
Total distance travelled
Total time taken
6.
Average speed =
7.
To find the distance travelled in the first n seconds:
Step 1: Let v = 0 to find t.
Step 2: Find s for each of the values of t found in step 1.
Step 3: Find s for t = 0 and t = n.
Step 4: Draw the path of the particle on a displacement-time graph.
Example 1
A particle moves in a straight line in such a way that, t seconds after passing
through a fixed point O, its displacement from O is s m. Given that s = 2 –
find
4
,
t+2
(i) expressions, in terms of t, for the velocity and acceleration of the particle,
(ii) the value of t when the velocity of the particle is 0.25 m s–1,
(iii) the acceleration of the particle when it is 1 m from O.
Solution
(i) s = 2 –
4
t+2
ds
4
=
(Apply the Chain Rule of Differentiation)
dt (t + 2)2
dv
8
a =
=–
dt
(t + 2)3
v =
(ii) When v = 0.25,
4
= 0.25
(t + 2)2
(t + 2)2 = 16
t+2=±4
t = 2 or t = – 6 (rejected)
(The negative value of t is rejected since
time cannot be negative.)
Kinematics
171
(iii) When s = 1,
4
2–
=1
t+2
4
=1
t+2
t+2=4
t=2
Substitute t = 2 into a = –
8
:
(t + 2)3
8
(2 + 2)3
1
=–
8
∴ Acceleration of the particle when it is 1 m from O is – 1 m s – 2.
8
a = –
Example 2
A particle moves in a straight line such that its displacement, s m from a fixed point
A, is given by s = 2t + 3 sin 2t, where t is the time in seconds after passing point A.
Find
(i) the initial position of the particle,
(ii) expressions for the velocity and acceleration of the particle in terms of t,
(iii) the time at which the particle first comes to rest.
Solution
(i) s = 2t + 3 sin 2t
When t = 0, s = 2(0) + 3 sin 2(0) = 0.
∴ The particle is initially at point A.
(ii) s = 2t + 3 sin 2t
ds
v =
dt
= 2 + 6 cos 2t
dv
a =
dt
= –12 sin 2t
172
UNIT 16
(iii) When v = 0,
2 + 6 cos 2t = 0
1
3
2t = 1.91 (to 3 s.f.)
t = 0.955
cos 2t = –
Example 3
A stone that was initially at rest was thrown from the ground into the air, rising at a
velocity of v = 40 – 10t, where t is the time taken in seconds.
(i) Find the maximum height reached by the stone.
(ii) Find the values of t when the particle is 35 m above the ground.
Solution
∫ v dt
= ∫ (40 – 10t) dt
(i) s =
= 40t – 5t2 + c
When t = 0, s = 0 ∴ c = 0
s = 40t – 5t2
At maximum height,
v = 0 (v = 0 at maximum displacement.)
40 – 10t = 0
t=4
When t = 4, s = 40(4) – 5(4)2 = 80.
∴ The maximum height reached by the stone is 80 m.
(ii) When s = 35,
40t – 5t2 = 35
2
5t – 40t + 35 = 0
t2 – 8t + 7 = 0
(t – 1)(t – 7) = 0
t = 1 or
t=7
∴ The particle is 35 m above the ground when t = 1 and t = 7.
Kinematics
173
Example 4
A particle moves in a straight line so that, t seconds after passing through a fixed
point O, its velocity, v cm s–1, is given by v = 8t – 3t2 + 3. The particle comes to
instantaneous rest at the point P. Find
(i) the value of t for which the particle is instantaneously at rest,
(ii) the acceleration of the particle at P,
(iii) the distance OP,
(iv) the total distance travelled in the time interval t = 0 to t = 4.
Solution
(i) When v = 0,
8t – 3t2 + 3 = 0
3t2 – 8t – 3 = 0
(3t + 1)(t – 3) = 0
t=–
1
(rejected) or
3
t=3
(ii) v = 8t – 3t2 + 3
a = 8 – 6t
When t = 3, a = –10
∴ Acceleration of the particle at P is –10 cm s–2.
∫ v dt
= ∫ (8t – 3t + 3) dt
(iii) s =
2
= 4t2 – t3 + 3t + c
When t = 0, s = 0 ∴ c = 0
∴ s = 4t2 – t3 + 3t
When t = 3, s = 18
∴ OP = 18 cm
174
UNIT 16
(iv) When t = 0, s = 0.
When t = 3, s = 18.
When t = 4, s = 12.
t=4
t=3
t=0
0
12
(Draw the path of
the particle on a
displacement-time graph.)
18
∴ Total distance travelled = 18 + (18 – 12)
= 24 cm
Example 5
A particle moving in a straight line passes a fixed point O with a velocity of 4 m s–1.
The acceleration of the particle, a m s–2, is given by a = 2t – 5, where t is the time
after passing O. Find
(i) the values of t when the particle is instantaneously at rest,
(ii) the displacement of the particle when t = 2.
Solution
∫ a dt
= ∫ (2t – 5) dt
(i) v =
= t2 – 5t + c
When t = 0, v = 4 ∴ c = 4
∴ v = t2 – 5t + 4
When the particle is instantaneously at rest, v = 0.
t2 – 5t + 4 = 0
(t – 4)(t – 1) = 0
t = 4 or
t=1
Kinematics
175
∫ v dt
= ∫ (t – 5t + 4) dt
(ii) s =
2
1 3 5 2
t – t + 4t + c1
3
2
When t = 0, s = 0 ∴ c1 = 0
1
5
∴ s = t 3 – t 2 + 4t
3
2
When t = 2,
1
5
s = (2)3 – (2)2 + 4(2)
3
2
2
=
3
2
∴ Displacement of the particle is m
3
=
Example 6
A particle starts at rest from a fixed point O and travels in a straight line so that,
t seconds after leaving point O on the line, its acceleration, a m s–2, is given by
a = 2 cos t – sin t. Find
(i) the value of t when the particle first comes to an instantaneous rest,
(ii) the distance travelled by the particle in the first 3 seconds after leaving O.
Solution
(i) a = 2 cos t – sin t
∫ a dt (Recall that when a particle is at instantaneous rest, v = 0.)
= ∫ (2 cos t – sin t) dt
v =
= 2 sin t + cos t + c
When t = 0, v = 0 ∴ c = –1 (Note that cos 0 = 1.)
∴ v = 2 sin t + cos t – 1
176
UNIT 16
When v = 0,
2 sin t + cos t – 1 = 0
2 sin t + cos t = 1 (R-formula is needed to solve this equation.)
5 sin (t + 0.4636) = 1
1
sin (t + 0.4636) =
5
basic angle, α = 0.4636 (to 4 s.f.)
t + 0.4636 = 0.4636, 2.677
t = 0, 2.21 (to 3 s.f.)
∴ The particle first comes to an instantaneous rest when t = 2.21.
∫ v dt
= ∫ (2 sin t + cos t – 1) dt
(ii) s =
= sin t – 2 cos t – t + d
When t = 0, s = 0 ∴ d = 2
∴ s = sin t – 2 cos t – t + 2
When t = 0, s = 0.
When t = 2.214, s = 1.785.
When t = 3, s = 1.121.
t=3
t = 2.214
t=0
0
1.121
1.785
∴ Total distance travelled = 1.785 + (1.785 – 1.121)
= 2.45 m (to 3 s.f.)
Kinematics
177
MATHEMATICAL FORMULAE
178
Mathematical Formulae